Lakhmir Singh Manjit Kaur Physics 2020 2021 for Class 10 Science Chapter 1 - Electricity

Answer:

The unit joule/coulomb is called by another name, volt. Volt is the SI unit of potential difference.
Potential difference = Work done/Quantity of charge moved

Answer:

(b) A volt is a joule per coulomb.
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to another.
1 volt = 1 joule / 1 coulomb

Answer:

(a) The acronym p.d. stands for the potential difference between two points in an electric circuit.
(b) Potential difference (p.d.) is measured by a device called voltmeter.

Answer:

When we say that the electric potential at a point is 1 volt, it means that 1 joule of work has been done in moving 1 coulomb of positive charge from infinity to that point.

Answer:

Charge given, Q = 1 C
Potential difference, V = 1 V
Thus, the amount of work done, W in moving the charge is:
W = V × Q
= 1 V × 1 C
= 1 J

Answer:

The SI unit of potential difference is volt and is denoted by the letter, V.

Answer:

The amount of charge, Q that flows between two points at a potential difference, V (= 12 V) is 2 C.
Therefore, the amount of work done, W in moving the charge is given by:

W = V × Q
= 12 V × 2 C
= 24 J

Answer:

The SI unit of electric charge is coulomb and is denoted by the letter, C.

Answer:

One coulomb is the quantity of electric charge which exerts a force of $9\times {10}^9\mathrm{N}$ on an equal charge placed at a distance of one metre from it.

Answer:

(a) Potential difference is measured in volts by using a voltmeter placed in parallel across a component.
(b) Copper is a good conductor. Plastic is an insulator.

Answer:

The substances through which electricity can flow are called conductors. Two examples are copper and aluminium.

The substances through which electricity cannot flow are called insulators. Two examples are glass and paper.

Answer:

Conductors: silver, copper, aluminium, nichrome, graphite, manganin, mercury
Insulators: cotton, air, paper, porcelain, mica, bakelite, polythene, sulphur

Answer:

The electric potential (or potential) at a point in an electric field is defined as the work done in moving a unit positive charge from infinity to that point. It is denoted by the symbol, V and its unit is volt.

Answer:

(a) The potential difference between two points in an electric circuit is equal to the amount of work done in moving a unit charge from one point to another point.
$\mathrm{Potential} \mathrm{difference} \left(\mathrm{V}\right)=\frac{\mathrm{Work} \mathrm{done} \left(\mathrm{W}\right)}{\mathrm{Quantity} \mathrm{of} \mathrm{charge} \mathrm{moved} \left(\mathrm{Q}\right)}$
(b) Here:
Potential difference, V = 230 − 220 = 10 V
Charge moved, Q = 4 C
So, using the relation V = W/Q, we have:
W = V $\times$ Q
= 10 $\times$ 4 = 40 J

Answer:

(a) The potential difference across a conductor is measured by an instrument called voltmeter.
(b) The term 'each coulomb' means 'every 1 coulomb'.
So, charge, Q = 1 C
Potential difference, V = 12 V
Now:
V = W/Q
12 = W/1
So, work done, W = 12 x 1 = 12 J
Since the work done on each coulomb of charge is 12 J, the energy transferred to each coulomb of charge is also 12 J.

Answer:

(a) The potential difference between two points in an electric circuit is equal to the amount of work done in moving a unit charge from one point to another point.
$\mathrm{Potential} \mathrm{difference} \left(V\right)=\frac{\mathrm{Work} \mathrm{done} \left(W\right)}{\mathrm{Quantity} \mathrm{of} \mathrm{charge} \mathrm{moved} \mathit{\left(}Q\right)}$
The SI unit of potential difference is volt.

(b) The potential difference between two points is said to be one volt if one joule of work is done in moving one coulomb of electric charge from one point to the other.
Thus:
1 volt = 1 joule/1 coulomb

(c) Here:
Work, W = 250 J
Charge, Q = 20 C
Using the relation, V = W/Q:
Potential difference, V = 250/20 = 12.5 V

(d) A voltmeter is an instrument that is used to measure the potential difference across a conductor.
A voltmeter is always connected in parallel across the points where the potential difference is to be measured.


Suppose we have a conductor AB, such as a resistance wire, and we want to measure the potential difference across its ends.
So one end of the voltmeter is connected to point A and the other end to the point B. We can read the value of potential difference in volts on the dial of the voltmeter.

(e) A voltmeter has a high resistance so that it takes a negligible current from the circuit.
In a circuit, a voltmeter is connected in parallel across the element of whose potential difference it has to measure. Since we don't want the current diverted through the voltmeter, the voltmeter has a high internal resistance so that the circuit maintains the same current as if the voltmeter was not present.

Answer:

 (b) potential difference                                                                                                                                                    
The work done in moving a unit charge across two points in an electric circuit is a measure of potential difference.

Answer:

d) voltmeter
It is connected in parallel across the points where the potential difference is to be measured.

Answer:

(d) coulomb
It is the SI unit of electric charge.

Answer:

(c) volt
It is the SI unit of potential difference.

Answer:

(d) 6.25 × 10¹⁸ electrons
Since the charge of an electron is $1.6\times {10}^{-19}\mathrm{C}$, 6.25 × 10¹⁸ electrons taken together constitute (6.25 × 10¹⁸ × 1.6 × 10⁻¹⁹ C) or 1 coulomb of charge.

Answer:

(a) The potential difference across each cell is 2 V. So, the potential difference at the terminals of the battery is 3 x 2 V = 6 V.

(b)
(i) For one cell:
V = 2 V
Q = 1 C
And also:
Electrical energy gained = Work done
So using the relation, V = W/Q or W = V x Q
= 2 x 1 = 2 J
(ii) For three cells:
V = 6 V
Q = 1 C
So, W = V x Q = 6 x 1 = 6 J
So 2 joules and 6 joules of electrical energy is gained for the above two cases (i) one cell and (ii) all three cells, respectively.

Answer:

In a copper atom, the electrons in the outermost orbit are lightly bound to the nucleus of the atom and so they move freely. This free flow of electrons makes copper a good conductor of electricity. In rubber, the electrons in the outermost orbits are tightly bound to the nucleus. So, they do not flow freely and hence, rubber is not a good conductor of electricity. 

Answer:

The electric current is the physical quantity that is expressed as coulomb/second since current, I = Q/t.

Answer:

The flow of electric charge in a conductor is called electric current. If a charge of Q coulombs flows through a conductor in time t seconds, the magnitude of electric current is given by, I = Q/t.

Answer:

When we switch on a light, electrons travel through the wires. The flow of these electrons is called electric current.

Answer:

The flow of negatively charged particles (electrons) in a metallic conductor constitutes electric current.

Answer:

(a) The direction of conventional current is from the positive terminal of a cell or battery to the negative terminal through the outer circuit.
(b) The flow of electrons is from the negative terminal to the positive terminal of a cell.

Answer:

Because current, I = Q/t, the correct relationship is $1 \mathrm{A}=\frac{1 \mathrm{C}}{1 \mathrm{s}}$.

Answer:

The SI unit of electric current is ampere and is denoted by the letter, A.

Answer:

(a) 1 ampere = 1000 milliamperes = ${10}^3 \mathrm{milliamperes}$
(b) 1 ampere = 1,000,000 microamperes = ${10}^6\mathrm{microamperes}$

Answer:

Ammeter.
Ammeter is connected in series with the circuit, whereas voltmeter is connected in parallel across the points where the p.d. is to be measured.

Answer:

The ammeter is connected in series with the circuit in which the current is to be measured, whereas the voltmeter is connected across the points where the potential difference is to be measured.

Answer:

(i) Variable resistance or rheostat. A rheostat is used to change the resistance in a circuit.
(ii) A closed switch. When the switch is closed, the circuit is completed and current flows through it.

Answer:

Here:
Charge, Q = 20 C
Time, t = 1 s
We know that, I = Q/t
So:
Current, I = 20/1 = 20 A
Thus, the electric current flowing through the circuit is 20 A.

Answer:

Here:
Current, I = 4 A
Time, t = 10 s
We know that I = Q/t.
So:
4 = Q/10
or Q = 4 x 10 = 40 C
Thus, the amount of electric charge that flows through the circuit is 40 coulombs.

Answer:

Here:
Charge, Q = 20 C
Time, t = 40 s
We know that, I = Q/t.
So:
I = 20/40 = 0.5 A
Therefore, the electric current flowing through the circuit is 0.5 A.

Answer:

(a) A current is a flow of electrons. For this to happen, there must be a closed circuit.
(b) Current is measured in amperes using an ammeter placed in series in a circuit.

Answer:

(a) A cell or a battery helps to maintain potential difference across a conductor.
(b) Here:
Potential difference, V = 10 V
Current, I = 2 A
Time, t = 1 minute = (1 x 60) s = 60 s
We know that the amount of energy transferred is equal to the work done.
Now:
V = W/Q
Or W = V x Q
= V x I x t (since Q = I x t)
= 10 x 2 x 60
= 1200 J
Thus, the amount of energy transferred is 1200 J.

Answer:

(a) An electric current is the flow of electric charge called electrons in a conductor and its magnitude is the amount of electric charge passing through a given point of the conductor in one second. The potential difference between the ends of the wire makes the current flow through the wire.

(b) The SI unit of electric current is ampere.
When one coulomb of charge flows through any cross-section of a conductor in one second, the electric current flowing through it is said to be one ampere.
1 ampere= $\frac{1 \mathrm{coulomb}}{1 \mathrm{second}}$

Answer:

An ammeter is an instrument that is used to measure electric current. It is always connected in series with the circuit in which the current is to be measured.

Suppose we want to find out the current flowing through a conductor AB. Here, we should connect the ammeter in series with the conductor AB as above.

Answer:

(a) If a charge of Q coulombs flows through a conductor in time t seconds, the magnitude, I of the electric current flowing through it is given by:
$I\mathit{=}\frac{\mathit{Q}}{\mathit{t}}$
(b) Given:
Current, I = 0.36 A
Time, t = 15 minutes = (15 $\times$ 60) s = 900 s
Using the formula, $I\mathit{=}\frac{\mathit{Q}}{\mathit{t}}$:
or Q = I $\times$ t = 0.36 $\times$ 900 = 324 C
Thus, the amount of electric charge that flows through the circuit is 324 C.

Answer:

(a) An ammeter should have a very low resistance so that it will not change the value of the current flowing in the circuit.
(b) A voltmeter should have a high resistance so that it can take only a negligible amount of current from the circuit.

Answer:

(a) Fixed resistance



(b) Variable resistance


(c) A cell

(d) A battery of three cells


(e) An open switch

(f) A closed switch

Answer:

A circuit diagram is a diagram that indicates how different components in a circuit have been connected, using electrical symbols for the components.
Here is the diagram of an electric circuit comprising a cell, a resistance, an ammeter, a voltmeter and a closed switch.

A voltmeter has a high resistance so that it takes only a negligible amount of current from the circuit.

Answer:

Given:
Current, I = 1 A
Time, t = 1 s
Using the formula:
$$\begin{gathered} I\mathit{=}\frac{\mathit{Q}}{\mathit{t}} \\ \mathrm{or}, \\ Q\mathit{=}I\mathit{\times }t \\ =1\times 1 \\ =1 \mathrm{C} \end{gathered}$$
Now, if the charge is $1.6\times {10}^{-19}\mathrm{C}$, the number of electrons is 1.
So, if the charge is 1 C, the number of electrons is given by:

$\frac{1}{1.6\times {10}^{-19}}\times 1=6.25\times {10}^{18}$
Thus, $6.25\times {10}^{18}$electrons should pass through a conductor in 1 second to constitute 1 ampere current.

Answer:

 Amount of energy = Work done
(a) Using the formula, V = W / Q, we have, W = V x Q
= 12 x 1 = 12 J

(b) W = V x Q = 12 x 5 = 60 J

(c) W = V x Q
= V x (I x t)
= 12 x 2 x 10 = 240 J

Answer:

Given:
Time, t = 10 s
Charge, Q = 25 C
Energy delivered = Work done, W = 200 J
(a) Using the relation, V = W/Q
V = 200/25 = 8 V
So the p.d. across the battery is 8 volts.
(b) Using the relation, I = Q/t,
I = 25/10 = 2.5 A
So, a current of 2.5 A flows from the battery.

Answer:

(a) Electric current is the flow of electric charges (called electrons) in a conductor such as a metal wire and its magnitude is the amount of electric charge passing through a given point of the conductor in one second.
i.e. I = Q/t
The SI unit of electric current is ampere.

(b) Here:
Charge, Q = 1 C
Time, t = 1 s
Then the current is given by:
I = Q/t
= 1/1 = 1 A

(c) Current is measured by an instrument called ammeter. An ammeter should be connected in series with the circuit in which the current is to be measured.

(d) The conventional direction of the flow of electric current is from the positive terminal of a cell or a battery to its negative terminal through the outer circuit, whereas the direction of flow of electrons is from the negative terminal to the positive terminal of a cell.

(e) Given:
Charge, Q = 10 C
Time, t = 0.01 s
Then the current, I = Q / t
So I = 10/0.01 = 1000 A
Also given that:
V = 10 mV = 1,00,00,000 V
So energy = Work done, W = V x Q
= 1,00,00,000 x 10 = 1,00,000,000 J = 100 MJ

Answer:

(c) voltage
The term voltage came from volt, which is the SI unit of potential difference.

Answer:

The correct statement is:
(d) 1
An ammeter is connected in series in a circuit and a voltmeter is connected in parallel. The other two statements are incorrect as an ammeter should have a very low resistance and a voltmeter should have a high resistance.

Answer:

(d) Ampere
It is the SI unit of electric current.

Answer:

(d) 50 C

Here, current, I = 5 A
Time, t = 10 s
So, using the relation, I = Q/t:
Q = I x t
Charge, Q = 5 x 10 = 50 C
Thus, a charge of 50 C is passed in 10 s.
Hence, the correct option is (d) 50 C.

Answer:

(c) 0.5 A

Here, charge, Q = 300 C
Time, t = 10 minutes = (10 x 60) s = 600 s
Then, current, I = Q / t
= 300 / 600 = 0.5 A
Hence, the correct option is (c) 0.5 A.

Answer:

(a) The lamps are connected in series.
(b) The student has connected the ammeter in parallel with the lamps. It should be connected in series.
(c) The correct diagram that shows how to connect the circuit:

Answer:

Here is the circuit diagram that shows how three bulbs can be lit from a battery so that two bulbs are controlled by the same switch. The third bulb has its own switch.

Answer:

Here, potential, V = 230 V
Current, I = 8 A
(a) The magnitude of the electric current is the amount of charge passing through a given point in one second. So, in this case, a charge of eight coulombs flows around the circuit each second.

(b) Energy transferred = Work done, W
Using the relation, V = W/Q:
W = V x Q
= 230 x 8 = 1840 J
Thus, 1840 joules of energy is transferred to the heater each second.

Answer:

Current, I = 5 A
Time, t = 1 s
Using the relation, I = Q / t:
Q = I x t
= 5 x 1 = 5 C
Now, if the charge is $1.6\times {10}^{-19}C$, the number of electrons is one.
So, if the charge is 5 C, the number of electrons is given by:
$\frac{1}{1.6\times {10}^{-19}}\times 5=31.25\times 10$¹⁸

Answer:

Ohm's law gives the relationship between the current in a conductor and the potential difference across its ends. According to this law, at constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.

Answer:

The SI unit of electrical resistance is ohm and its symbol is $\mathrm{\Omega }$.

Answer:

Electrical resistance is the physical quantity whose SI unit is ohm.

Answer:

Objects that have infinitely high electrical resistance are called insulators. An insulator does not allow electricity to flow through it.

Answer:

From Ohm's law we have:
V / I = R
Or I = V/R
Now, according to the question, the resistance is constant and the potential difference is halved.
i.e. R' = R
and V' = V/2
Then the new current is given by:
I' = V'/R'
= V/2/R
= (1/2) (V/R)
= (1/2) I
It is clear from the above statement that the current becomes half.

Answer:

The strength of the electric current flowing in a given conductor depends on two factors:
1) The potential difference across the ends of the conductor
2) The resistance of the conductor

Answer:

A thick wire has a greater area of cross-section, whereas a thin wire has a smaller area of cross-section. Also, the resistance of a conductor is inversely proportional to its area of cross-section. This means that a thick wire has lesser resistance than a thin wire.

Answer:

From Ohm's law we have:
V/I = R
Or I = V/R
Now, according to the question, the resistance of the circuit is halved and the potential difference is constant.
i.e. R' = R/2
and V' = V
Then the new current is given by:
I' = V'/R'
= V/R/2
= 2V/R
= 2I
It is clear from the above statement that the current becomes double.

Answer:

Here:
Potential difference, V = 20 V
Resistance, R = 5 Ω
Substituting these values in Ohm's equation, V/I = R:
Current, I = V/R = 20/5 = 4 A
Thus, the current flowing in the circuit is 4 A.

Answer:

Here:
Resistance, R = 20 Ω
Current, I = 2 A
Substituting these values in the Ohm's equation, V/I = R:
Potential difference, V = I x R = 2 x 20 = 40 V

Answer:

Here:
Current, I = 5 A
Potential difference, V = 3 V
Substituting these values in the Ohm's equation, V/I = R:
Resistance of the wire, R = 3/5 = 0.6 Ω

Answer:

Ohm's law state a relation between potential difference and current.

Answer:

Substances that conduct electricity easily and offer very low resistance (opposition) to the flow of current are called good conductors. Examples: copper, silver

A resistor is an electrical component that limits or regulates the flow of electrical current in an electronic circuit. Resistors have comparatively high electrical resistance. Examples: constanan, nichrome

The substances that have infinitely high electrical resistance are called insulators. Examples: wood, rubber

Answer:

Conductors: mercury, aluminium, iron and metal coin

Resistors: nichrome and manganin

Insulators: rubber, polythene, wood, paper, thermocol and bakelite

Answer:

According to Ohm's law, the current (I) flowing through a wire is directly proportional to the potential difference (V) across it, provided its temperature remains the same.
Thus I α V.
V / I = Constant = R
V = IR
Here R is the resistance.
The unit of resistance is ohm (Ω).

One ohm (Ω) is the resistance of a conductor when a potential difference of one volt is applied to its ends and a current of one ampere flows through it.

Answer:

(a) Resistance is the property of a conductor to resist the flow of charge through it. The relation between resistance, current and potential difference is V = IR.
(b) Voltage, V = 12 V
    Resistance, R = ?
    Current, I = 2.5 $\times {10}^{-3}$ A
    V = IR
    R = V/I
    R = 12/2.5 $\times {10}^{-3}$
    R = 4800 Ώ

Answer:

​(a) One ohm (Ω) is the resistance of a conductor when a potential difference of one volt is applied to its ends and a current of one ampere flows through it.

(b) The resistance of a wire can be represented by the following equation:

 $R=\frac{\mathrm{\rho }l}{A}$

So, when the conductor is made thinner, its area of cross-section will decrease, since resistance is inversely proportional to the area of the conductor. Therefore, its resistance will increase.

(c) We know that V = IR

If V is constant, resistance is doubled  R' = 2R.

Then V' = I'R'

V' = V since V is constant

$V=I\text{'}\times 2\times R$

$I\text{'}=\frac{V}{2R}=\frac{I}{2}$

Thus, the current will get reduced to half.

Answer:

(a) Electricians wear rubber hand gloves while working with electricity because rubber is an insulator and it will prevent them from getting electric shock, in case of short circuit or overloading.

(b) We know that V = IR

Here, I = 6 A

R = 40 Ω

So, V = 6 x 40

V = 240 V

So 240 V is the potential difference needed to send a current of 6 A through an electrical appliance that has a resistance of 40 Ω.

Answer:

(i)


(ii) When potential difference = 0.8 V, current = 0.32 A.
      $$\begin{gathered} \frac{\mathrm{V}}{\mathrm{I}}=\frac{0.8}{0.32} \\ \frac{\mathrm{V}}{\mathrm{I}}=2.5 \end{gathered}$$
When potential difference = 1.2 V, current = 0.48 A.
      $$\begin{gathered} \frac{\mathrm{V}}{\mathrm{I}}=\frac{1.2}{0.48} \\ \frac{\mathrm{V}}{\mathrm{I}}=2.5 \end{gathered}$$
When potential difference = 1.6 V, current = 0.64 A.
      $$\begin{gathered} \frac{\mathrm{V}}{\mathrm{I}}=\frac{1.6}{0.64} \\ \frac{\mathrm{V}}{\mathrm{I}}=2.5 \end{gathered}$$
The V-I graph is a straight line with a constant slope of 2.5.

(iii)  The resistance of the wire is 2.5 Ω.

Answer:

 (a) The ratio of potential difference and current is known as resistance R.
(b)

From the graph, resistance, R =$\frac{V}{I}=\frac{5}{0.2}=25 \mathrm{\Omega }$

(c) Ohm's law is illustrated by the above V−I graph.

Answer:

 (c) 2 A

If the p.d. across a 3 Ω resistor is 6 V, the current flowing in the resistor will be 2 A as current (I) is given by the equation, I = V (Voltage) / R (Resistance).

or  I = V/R

I = 6 V/3 Ω

I = 2 A

Answer:

(d) 24 Ω

Voltage, V = 12 V
Current, I = 0.5 A
Resistance, R = Voltage/Current
R = V/I
R = 12/0.5
R = 24 Ω
Thus, if a car's headlight bulb working on a 12 V car battery draws a current of 0.5 A, the resistance of the light bulb will be 24 Ω.

Answer:

 (c) 9.2 A

Resistance = 25 Ω

Voltage = 230 V

Current = Voltage/Resistance

I = V/R

I = 230/25

I = 9.2 A

Thus, if an electrical appliance has a resistance of 25 Ω and when that electrical appliance is connected to a 230 V supply line, the current passing through it will be 9.2 A.

Answer:

 (b) 3
The number of coulombs passing through the resistor is the current passing through it.
Current = Voltage/Resistance
I = V/R
I = 12/4
I = 3 A
Thus, when a 4 Ω resistor is connected across the terminals of a 12 V battery, the number of coulombs passing through the resistor per second will be 3.

Answer:

(d) current and potential difference
Ohm's law gives the relationship between current and potential difference.

Answer:

(d) ohm
The unit of electrical resistance is ohm.

Answer:

 (d) insulator
A substance having infinitely high electrical resistance is called an insulator.

Answer:

 (b) half

As we know from Ohm’s law:
Voltage = Current x Resistance
V = IR
If the voltage is constant, the resistance is doubled and the current becomes half.

Answer:

 (d) double

As we know from Ohm’s law:
Voltage = Current x Resistance
V = IR
If the voltage is constant, the resistance of the circuit is halved. That is, it becomes R / 2.
Current, I = V/R
I = V/R
I = V/R/2 = 2 I
Thus by keeping the p.d. constant, the resistance of a circuit is halved and the current is doubled.

Answer:

An electric room heater draws a current of 2.4 A from the 120 V supply line.

So current, I = 2.4 A

Voltage, V = 120 V

Resistance, R = V/I

R = 120/2.4 = 50 Ω

The resistance will remain constant unless the room heater is changed.

Now when the room heater is connected to the 240 V supply line, the current drawn is:

I = V/R

I = 240/50 = 4.8 A

So the room heater will draw a 4.8 A current when connected to the 240 V supply line.

Answer:

 Resistance is represented by the symbol Omega, Ω. Resistance has the following electrical properties:

1. The resistance of a conductor depends on its length. It is directly proportional to the length, R α L.
2. The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A.
3. Resistance depends on the nature of the material of the conductor.
4. Resistance also depends on the temperature of the conductor.

Answer:

 (a) Ohm's law states that the graph between V and I for a conductor is a straight line passing through the origin.

(b) Ohm's law is true only if the temperature of the conductor is constant.

Answer:

A potential difference of 10 V is needed to make a current of 0.02 A flow through a wire.
So potential difference, V = 10 V
Current, I = 0.02 A
Resistance of the wire, R = V/I
R = 10/0.02 = 500 Ω
The resistance will remain the same unless the wire is changed.
Now to make a current of 250 mA to flow through the wire, the potential difference required is:
V = IR
V = 250 x 10^-3 x 500
V = 125 V
So a potential difference of 125 V is needed to make a current of 250 mA flow through the wire.

Answer:

A current of 200 mA flows through a 4 kΩ resistor.
So current, I = 200 mA = 200 x 10^-3A
Resistance, R = 4 kΩ = 4000 Ω
Potential difference, V = IR
V = 200 x 10^-3 x 4000
V = 800 V
So the potential difference across the resistor will be 800 V.

Answer:

The resistance of a conductor is inversely proportional to its area of cross-section, i.e.R α 1 / A. So when the conductor is made thicker, its resistance decreases.

Answer:

Answer:

 The resistance of a conductor depends on the following factors:

1. The resistance of a conductor depends on its length. It is directly proportional to the length, R α l.
2. The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A.
3. Resistance depends on the nature of the material of the conductor.
4. Resistance also depends on the temperature of the conductor.

Answer:

Silver is the best conductor of electricity.

Answer:

Iron is a better conductor of electricity than mercury because the resistivity of iron is less than that of mercury.

Answer:

Copper and aluminium wires are used for electrical transmission because they have a low resistance, which makes them good conductors of electricity.

Answer:

Answer:

Nichrome is an alloy of nickel, chromium, manganese and iron. It is used to make the heating elements of electrical appliances because it has a high resistance and a high melting point. 

Answer:

Nichrome is used to make the heating elements of electrical appliances because it has a high resistance and a high melting point.

Answer:

Answer:

Answer:

Answer:

Resistance is measured in ohm (Ω). The resistance of a wire increases as the length increases; as the temperature increases; and as the cross-sectional area decreases.

Answer:

​

(a) We know that resistance,  $R=\frac{\mathrm{\rho }l}{A}$

So resistivity,  $\mathrm{\rho }=\frac{RA}{l}$

When the area of cross-section, A is 1 m² and the length of the wire, l is 1 m, resistance is equal to resistivity.

Resistivity may be defined as the resistance offered by a conductor of length 1 m and area of cross-section 1 m².

(b) Given length of the wire, l = 1 m

Diameter of the wire, d = 0.2 mm

So the radius of the wire, r = 0.1 mm = 0.0001 m

Area of cross-section of the wire =  $A=\mathrm{\pi }{r}^2=3.14\times {\left(0.0001\right)}^2=3.14\times {10}^{-8} {\mathrm{m}}^2$

Resistance of the wire, R = 10 Ω

 So resistivity = $\mathrm{\rho }=\frac{RA}{l}$

 $\mathrm{\rho }=3.14\times {10}^{-7} \mathrm{\Omega } \mathrm{m}$

Thus, the resistivity of the material, $\mathrm{\rho }=3.14\times {10}^{-7} \mathrm{\Omega } \mathrm{m}$​.

Answer:

(a) Length of the cable wire = l 

Then,  $R=\frac{\mathrm{\rho }l}{A}$

$R=\frac{\mathrm{\rho }l}{A}$

Answer:

(a) Copper and aluminium are good conductors of electricity because their resistivity is very low.

(b) Length of the cable wire, l = 1 km = 1000 m

To find the resistance of the aluminium cable, R:

We know that $R=\frac{\mathrm{\rho }l}{A}$

$R=\frac{1.7\times {10}^{-8}\times 1000}{19.6\times {10}^{-8}}$

So, R = 86.5 Ω

Thus, the resistance of the copper wire, R = 86.5 Ω

Answer:

Answer:

1. The resistance of a conductor depends on its length. It is directly proportional to the length, R α l.
2. The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A.
3. The resistance also depends on the temperature of the conductor. It increases on raising the temperature and decreases on lowering the temperature.

Answer:

Let the resistance of the aluminium cable be R.

We know that $R=\frac{\mathrm{\rho }l}{A}$

$R=\frac{27\times {10}^{-8}\times 10000}{3.14\times {10}^{-8}}$

R = 86 Ω

Answer:

Answer:

Answer:

Resistance, 
$R=\frac{\mathrm{\rho }l}{A}$
 where l = length of  wire
            A= area of cross-section of the wire

(i) When the length is tripled, l' = 3l

New resistance,
$R\text{'}=\frac{\mathrm{\rho }l\text{'}}{A}$

$R\text{'}=\frac{\mathrm{\rho }\left(3l\right)}{A}=\frac{3\mathrm{\rho }l}{A}$

$R\text{'}=3R$

Thus, the resistance will become three times the original resistance.

(ii) When the diameter is tripled, d' = 3d

Radius, r' = 3r

Area of cross-section, $A\text{'}=\mathrm{\pi }r{\text{'}}^2=\mathrm{\pi }(3r{)}^2=9\mathrm{\pi }{r}^2=9A$

Thus, the area of cross-section will become nine times.

Then the new resistance,
$R\text{'}=\frac{\mathrm{\rho }l}{A}$

$R\text{'}=\frac{\mathrm{\rho }l}{9A}$

$R\text{'}=\frac{R}{9}$

Thus, the resistance will be reduced by nine times.

(iii) When the material is changed to one whose resistivity is three times:

Answer:

Answer:

(a) We know that resistance, 
$R=\frac{\mathrm{\rho }l}{A}$

So resistivity, 

$\mathrm{\rho }=\frac{RA}{l}$
When the area of cross-section, A = 1 m² and the length of the wire, l = 1 m, $\mathrm{\rho }=R$.

Resistivity is the resistance of the conductor whose area of cross-section, A is 1 m² and length, l is 1 m.

(d) Resistivity of a substance depends on the material and the temperature of the substance. It does not depend on the dimensions of the substance like length, radius etc.

(e) Length of the wire, l = 1 m

Diameter of the wire, d = 0.3 mm

So the radius of the wire, r = 0.15 mm = 0.00015 m = 1.5 x 10^-4 m

Area of cross-section of the wire,  $A={\mathrm{\pi r}}^2=3.14\times ( 1.5\times {10}^{-4} {)}^2=7.065\times {10}^{-8} {\mathrm{m}}^2$

Resistance of the wire, R = 26 Ω

So resistivity,
    $\rho =\frac{RA}{l}$

$$\begin{gathered} p=\frac{26\times 7.065\times {10}^{-8}}{1} \\ p=1.84 \times {10}^{-8} \mathrm{\Omega } \mathrm{m} \end{gathered}$$

Thus, the resistivity of the material is $1.84\times {10}^{-8 }\mathrm{\Omega } \mathrm{m}$.

Answer:

(d) 30 Ω

Resistance, $R=\frac{\mathrm{\rho }l}{A}$

Length, l = 300 m

Cross-section area, A = 1.0 mm² = 10^-6 m²

Resistivity, $\mathrm{\rho }$ = 1.0 × 10⁻⁷ Ωm

Resistance, $R=\frac{{10}^{-7}\times 300}{{10}^{-6}}$

R = 30 Ω

Answer:

(d) one-fourth
R = ρ $\frac{\mathit{l}}{\mathit{A}}$

When the diameter is doubled, d' = 2d

Radius, r' = 2r

Area of cross-section, $A\text{'}=\mathrm{\pi }r{\text{'}}^2=\mathrm{\pi }(2r{)}^2=4\mathrm{\pi }{r}^2=4A$

The area of cross-section will increase by four times.

Then the new resistance,$R\text{'}=\frac{\mathrm{\rho }l}{A\text{'}}$

$R\text{'}=\frac{\mathrm{\rho }l}{4A}$

$R\text{'}=\frac{R}{4}$

Thus, the resistance will get reduced by four times.

Answer:

(c) 60 Ω

If the resistance of a certain copper wire is 1 Ω, the resistance of a similar nichrome wire will be about 60 Ω because the resistivity of nichrome is 60 times the resistivity of copper.

Answer:

(a) four times
 Resistance of the wire is given by:
R = ρ $\frac{\mathit{l}}{\mathit{A}}$
When the diameter is halved:
$d\text{'}=\frac{d}{2}$

Radius:
$r\text{'}=\frac{r}{2}$

Area of cross-section:
$A\text{'}=\mathrm{\pi }r{\text{'}}^2=\mathrm{\pi }{\left(\frac{r}{2}\right)}^2=\frac{\mathrm{\pi }{r}^2}{4}=\frac{A}{4}$

Area of cross-section will get reduced by four times.

Then the new resistance:
$R\text{'}=\frac{\mathrm{\rho }l}{A\text{'}}$

$R\text{'}=\frac{4\mathrm{\rho }l}{A}$

$R\text{'}=4R$

Thus, the resistance will increase by four times.

Answer:

 (d) a semiconductor
The resistivity of a certain material is 0.6 Ωm. The material is most likely to be a semiconductor because it has moderate resistivity.

Answer:

 (b) half
We know that the resistance of a conductor is given by:
$\mathrm{R}=\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}$
where ρ = resistivity
l = length of the conductor
A = area of the cross-section of the conductor
Let the new resistance be R' when the area of cross-section of the conductor is doubled
${\mathrm{R}}^{,}=\mathrm{\rho }\frac{\mathrm{l}}{2\mathrm{A}}$

Thus, the new resistance becomes half of the previous one.

Answer:

 (c) temperature
The resistivity of copper depends only on temperature.

Answer:

(b) 2 times
The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A. So, when the area of cross-section of a resistance wire is halved, its resistance will increase by two times.

Answer:

We know that the resistance of a conductor is given by:
$\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}=20 \mathrm{\Omega }$
where ρ = resistivity
l = length of the conductor
A = area of cross-section of the conductor
Now the length is increased to twice the original length. Then let the new resistance be denoted by R'.

$$\begin{gathered} {\mathrm{R}}^{,}=\mathrm{\rho }\frac{2\mathrm{l}}{{\displaystyle \frac{\mathrm{A}}{2}}}=4\mathrm{\rho }\frac{\mathrm{l}}{{\displaystyle \mathrm{A}}} \\ {\mathrm{R}}^{,}=4\mathrm{R}=4\times 20=80 \mathrm{\Omega } \end{gathered}$$

Thus, the new resistance will become four times.

Answer:

(a) electric wires: They are made of conductors. Conductors have a low resistivity and so, we will use material Q. 
 

Answer:

The electrical resistivities of the four materials A, B, C and D are given. We can classify them as follows:

Answer:

(a) E is the best conductor of electricity because its resistivity is the least among the given values.
(b) C is a better conductor than A because its resistivity is less than that of A.
(c) B should be used for making heating elements of electric iron because its resistivity is the highest among the given values.
(d) E and C should be used for making electric wires because their resistivity is less.

Answer:

Answer:

If five resistors, each of value 0.2 Ω, are connected in series, the resultant resistance will be the sum of the individual resistances.
R = R₁ + R₂ + R₃ + R₄ + R₅
R = 0.2 + 0.2 +0.2 + 0.2 + 0.2
R = 5 x 0.2
R = 1 Ω

Answer:

In a parallel combination, the voltage remains the same across each resistor but the current gets divided. Let I_1, I₂ and I₃ be the current across resistors R_1, R₂ and R₃, respectively.

I = I₁ + I₂ + I₃

V = IR, V = I₁R₁, V = I₂R₂, V = I₃R₃

$$\begin{gathered} I=\frac{V}{R}, I_1=\frac{V}{R_1}, I_2=\frac{V}{R_2}, I_3=\frac{V}{R_3} \\ \mathrm{Using} \mathrm{these} \mathrm{values}, \mathrm{we} \mathrm{get}: \\ \frac{V}{R}=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3} \\ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \end{gathered}$$

So, the total resistance is the sum of the reciprocal of the individual resistances.

Answer:

If three resistors, each of 3 Ω, are connected in parallel combination, the total resistance of the parallel combination will be:

$$\begin{gathered} \frac{1}{R}=\frac{1}{R_{\mathit{1}}}+\frac{1}{R_{\mathit{2}}}+\frac{1}{R_{\mathit{3}}} \\ \mathit{ }\frac{1}{R}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3} \\ \mathrm{or} \frac{1}{R}=\frac{3}{3}=1 \mathrm{\Omega } \\ \mathrm{So} R=1 \mathrm{\Omega } \end{gathered}$$

The total resistance of the parallel combination is 1 Ω.

Answer:

 The two resistors of 2 Ω each should be connected in parallel combination to produce equivalent resistance of 1 Ω.

$$\begin{gathered} \mathrm{Resistance} \mathrm{in} \mathrm{parallel} \mathrm{arrangement} \mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{1}{R}=\frac{1}{R_{\mathit{1}}}+\frac{1}{R_2} \\ \frac{1}{R}=\frac{1}{2}+\frac{1}{2} \\ \mathrm{or} \frac{1}{R}=\frac{2}{2} \\ \mathit{ }R=1 \mathrm{\Omega } \end{gathered}$$

  R = 1

Total resistance, R = 1 Ω

Answer:

When two resistors, X and Y are connected in parallel combination, the total resistance will be less than the individual resistance. The total resistance in a parallel arrangement is given by:
 $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$
where as in a series combination, the total resistance is more than the individual resistances. The total resistance in a series combination is given by:
R = R₁+ R₂

Answer:

Case 1
Connecting a 2 Ω and 6 Ω resistor in series combination:
R = R₁ + R₂
R =2 $\mathrm{\Omega }$ + 6 $\mathrm{\Omega }$ =8 $\mathrm{\Omega }$
Thus, we will get an 8 Ω resistor.

Case 2
Connecting a 2 Ω and 6 Ω resistor in parallel combination:

$$\begin{gathered} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R}=\frac{1}{2}+\frac{1}{6} \\ \Rightarrow \frac{1}{R}=\frac{4}{6} \\ \Rightarrow R=\frac{6}{4} \mathrm{\Omega } \end{gathered}$$
Thus, we will get a 1.5 Ω resistor.

Answer:

(a) Case 1
Connecting two 4 Ω resistors in parallel combination:
$$\begin{gathered} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R}=\frac{1}{4}+\frac{1}{4} \\ \frac{1}{R}=\frac{2}{4} \\ \frac{1}{R}=\frac{1}{2} \\ R=2 \mathrm{\Omega } \\ \mathrm{Thus} \mathrm{we} \mathrm{get} \mathrm{a} 2 \mathrm{\Omega } resistor. \\ \mathrm{We} \mathrm{should} \mathrm{connect} \mathrm{two} 4 \mathrm{\Omega } \mathrm{in} \mathrm{parallel} \mathrm{to} \mathrm{get} \mathrm{a} \mathrm{resistance} \mathrm{of} 2 \mathrm{\Omega }. \end{gathered}$$

(b) Case 2
Connecting two 4 Ω resistors in series combination:
R = R₁ + R₂
R = 4 $\mathrm{\Omega }$ + 4 $\mathrm{\Omega }$ = 8 $\mathrm{\Omega }$
Thus, we will get an 8 Ω resistor.
We should connect two 4 $\mathrm{\Omega }$ resistors in parallel to get a resistance of 8 $\mathrm{\Omega }$.

 

Answer:

In figure A, the resistance, R_A = 10 Ω

In figure B, the resistors of 10 Ω and 1000 Ω are connected in parallel combination.
Therefore:

$$\begin{gathered} \frac{1}{\mathrm{R}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2} \\ \frac{1}{\mathrm{R}}=\frac{1}{10}+\frac{1}{1000} \\ \mathrm{or} \frac{1}{\mathrm{R}}=\frac{100}{1000}+\frac{1}{1000} \\ \mathrm{or} \frac{1}{\mathrm{R}}=\frac{101}{1000} \\ \mathrm{R}=\frac{1000}{101}=9.9 \mathrm{\Omega } \end{gathered}$$
R_B = 9.9 Ω
So, the arrangement B has lower resistance.

 

Answer:

A wire that has a resistance, R is cut into two equal pieces.
The resistance of the each piece will be  $\frac{\mathrm{R}}{2}$.

The two parts, each of resistance, $\frac{\mathrm{R}}{2}$ are joined in parallel.

The resistance of the parallel combination:
$$\begin{gathered} \frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2} \\ Here R_1=\frac{R}{2} \\ R_2=\frac{R}{2} \\ \frac{1}{R_P}=\frac{1}{{\displaystyle \frac{R}{2}}}+\frac{1}{{\displaystyle \frac{R}{2}}} \\ \frac{1}{R_P}=\frac{2}{R}+\frac{2}{R} \\ \frac{1}{R_P}=\frac{4}{R} \\ {R}_P =\frac{R}{4} \end{gathered}$$            
Thus, the resistance of the combination will be R/4.

Answer:

Case 1
Let the total resistance of the combination shown in the diagram below be R.
Here 500 Ω and 1000 Ω resistors are connected in series combination. Their net resistance is given by R = R₁+ R₂
Here:
R₁=500 Ω
R₂=1000 Ω
So:
R = 500 + 1000 = 1500 Ω
Thus, the total resistance of the combination is 1500 Ω.
 
Case 2
In the circuit diagram shown above, two 2 Ω resistors are connected in parallel combination.
Therefore:
$$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2} \\ \frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{1}{2}+\frac{1}{2} \\ \frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{2}{2} =1 \mathrm{\Omega } \end{gathered}$$
Thus the total resistance of the combination is 1 Ω.

Case 3
In the circuit diagram shown above, two 4 Ω resistors are connected in parallel combination.
$$\begin{gathered} Therefore their resul\tan t resis\tan ce is given by \\ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ Here R_1=4 \Omega \\ R_2=4 \Omega \\ so \\ \frac{1}{R}=\frac{1}{4}+\frac{1}{4} \\ or\frac{1}{R}=\frac{2}{4} \\ or R=2 \Omega \end{gathered}$$

In the above circuit diagram, this 2 Ω and 3 Ω resistors are connected in series combination. So, the total resistance of the circuit can be calculated as:
R = R₁ + R₂
here
R₁= 2 Ω
R₂= 3 Ω
So:
R = 2 Ω + 3 Ω = 5 Ω
Thus, the total resistance of the combination is 5 Ω.
 

Answer:

In the given circuit, two 6 Ω and 4 Ω resistors are connected in parallel combination across a cell of 24 V. We know that in a parallel combination, the voltage across the resistor remains the same and only the current gets divided. So the current (I₁) through the 6 Ω resistor is:
$$\begin{gathered} I_1=\frac{V}{R_1} \\ \mathrm{here} R_1=6 \mathrm{\Omega } \\ V=24 \mathrm{V} \\ \mathrm{so} \\ I_1 =\frac{24}{6} \\ \mathrm{or} I_1=4 \mathrm{A} \end{gathered}$$
The current (I₂) through the 4 Ω resistor is:
$$\begin{gathered} I_{\mathit{2}}=\frac{V}{R_{\mathit{2}}} \\ {R}_{\mathit{2}}=4 \mathrm{\Omega } \\ I=\frac{24}{4} \\ \mathrm{or}\mathit{ }I=6 \mathrm{A} \end{gathered}$$

Answer:

The series resistance is given by R = R₁ + R₂ + R₃ + R₄. Hence, it is more than the individual resistance.
The parallel resistance is given by 1 / R = 1 / R₁ + 1 / R₂ + 1 / R₃ + 1 / R₄. Hence, it is less than the individual resistance.

Answer:

When five resistors are connected in series, the total resistance is given by:                     
R = R₁+ R₂+ R₃+ R₄+ R₅
Here,
R₁= 0.2 $\mathrm{\Omega }$,
R₂= 0.3 $\mathrm{\Omega }$,
R₃= 0.4 $\mathrm{\Omega }$,
R₄= 0.5 $\mathrm{\Omega }$,
R₅= 0.12 $\mathrm{\Omega }$

Therefore:
R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω 
R = 13.4 Ω
Total resistance of the circuit = 13.4 Ω
The current flowing through this series combination is given by I = V / R.
or I = 9 / 13.4 = 0.67 A
Now since the resistors are connected in series, the current flowing through each resistance is the same. Hence, the current through the 12 Ω resistor is equal to 0.67 A.

Answer:

(a) Total resistance, R = 20 Ω + 4 Ω = 24 Ω
(b) Current through the circuit,
     I = V/R 
     I = 6/24
     I = 0.25 A
(c) Potential difference across the electric bulb,
 V_bulb = IR_bulb
          =0.25$\times$20
          = 5 V
(d) Potential difference across the resistance wire,
V_wire = IR_wire
         = 0.25$\times$4
         =  1 V

Answer:

(i) The resistance between B and C are in parallel. Hence the total resistance, R between B and C is given by:
 R_BC= [15 x 10 ] / [ 15 + 10 ] = 6 Ω ... (a)
Potential drop across B and C = IR = 1 x 6 = 6 V
Current through the 10 Ω resistor = V / R = 6 / 10 = 0.6 A
Current through the 15 Ω resistor = V / R = 6 / 15 = 0.4 A
(ii) p.d. across AB:
V_AB = 5 x 1 = 5 V
p.d. across AC:
V_AC = I x R_AC
R_AC = R + 5 = 11 Ω (see equation (a) )
 V_AC = 1 x 11 = 11 V
(iii) Total resistance, R_AC = R_BC + R_AC
                                      = 6 + 5
                                      = 11 Ώ

Answer:

 (1) The resistors of 6 Ω and 3 Ω are connected in series. Therefore, their net resistance can be calculated as:
  R = R₁ + R₂
Here, R₁ = 6 Ω
         R₂ = 3 Ω
 So:
R = 6 Ω + 3 Ω = 9 Ω
The current through this branch, I = V/R
I = 4/9 = 0.44 A
In a series combination, the current remains the same. So the current through the 6 Ω resistor is 0.44 A.

(2) The current through the branch with resistors of 12 Ω and 3 Ω:
I = V/R
I = 4/(12 + 3) = 4 / 15 A
The potential difference across the 12 Ω resistance can be obtained by using the equation,
V = IR.
V = (4 / 15) x 12 = 3.2 V

Answer:

(a) For the minimum current flowing in the circuit, the resistors should be connected in series and for the maximum current in the circuit, the resistors should be connected in parallel with the battery.
(b) When the resistors are connected in parallel:

$$\begin{gathered} \mathrm{Resistance} \mathrm{in} \mathrm{parallel} \mathrm{arrangement} \mathrm{is} \mathrm{given} \\ \frac{1}{R}=\frac{1}{R_{\mathit{1}}}+\frac{1}{R_{\mathit{2}}} \\ \mathrm{Here} R_1=5 \mathrm{\Omega } \\ \mathit{ }{R}_{\mathit{2}}=10 \mathrm{\Omega } \\ \frac{1}{R}=\frac{1}{5}+\frac{1}{10} \\ \mathrm{or} \frac{1}{R}=\frac{10+5}{50} \\ \mathrm{or} \frac{1}{ R}=\frac{15}{50} \\ \mathit{ }R=\frac{50}{15}=3.33 \mathrm{\Omega } \end{gathered}$$
Total resistance = 3.33 Ω
Therefore, strength of the total current, I = V / R
I = 6 / 3.33
I = 1.8 A
When the resistors are connected in series, the resultant resistance is given by R = R₁ + R₂
Here, R₁ = 5 $\mathrm{\Omega }$
          R₂= 10 $\mathrm{\Omega }$
    So, R = 5 $\mathrm{\Omega }$ + 10 $\mathrm{\Omega }$ = 15 $\mathrm{\Omega }$
Total resistance = 15 Ω
Therefore, strength of the total current, I = 6 / 15 = 0.4 A

Answer:

(i) As shown in the figure, the resistors R₂ and R3 are connected in parallel. Their total resistance is given by:

1/R = 1/R₂+ 1/R₃

Here, R₂ = 3 Ω 
         R₃ = 6 Ω 
So,
     1/R = 1/3 + 1/6
Or 1/R  = (2+1)/6     
1/R = 3/6     
R = 2 Ω 
This resistance is in series with the resistor, R₁. 
Total resistance = 2 Ω + R₁       
R₁ = 4 Ω
Therefore, total resistance = 2 Ω + 4 Ω = 6 Ω           


(ii) The current through the circuit can be calculated as follows:
Current, I = V / R
I = (12 / 6) A
I = 2 A

(iii) The potential difference across R₁ = 2 A x 4 Ω = 8 V

Answer:

Given I = 1 A (Across 5 Ω)
         R = 5 Ω
The potential drop across AB, V = I $\times$ R
                                       or  V = 5 Ω x 1 A = 5 V
In a parallel circuit, the potential difference across the ends of all resistors remains the same. Therefore, the potential difference across 4 Ω and 10 Ω will be 5 V:
The current flowing through the 4 Ω resistor, I = V/R.
Here, V = 5 V
         R = 4 Ω
So,  I = 5/4 = 1.25 A
The current flowing through the 10 Ω resistor, I = V/R.
Here, V = 5 V
         R = 10 Ω                                                              
So.   I = 5/10 = 0.5 A

Answer:

Suppose x resistors should be connected in parallel to draw a current of 5 A.
For a parallel combination, 1 / R = 1 / R₁ + 1 / R₂ + 1 / R₃ + ... (x times)

Here, R₁ = R₂ = R₃ = ... = 176 Ω
Then the resultant resistance, R = 176 / x Ω

Given that the current, I = 5 A and voltage, V = 220 V

Then by Ohm's law, V = IR

By substituting the values, we get:

x = 4

Hence, four resistors of 176 Ω each should be connected in parallel to draw a current of 5 A from a 220 V supply line.

Answer:

(a) When only one coil, A is used:
V = IR
220 = 24 I
I = 9.2 A

(b) When coils, A and B are used in series:
Total resistance R = R_A+ R_B = 48 Ω
V = IR
220 = 48 I
I = 4.58 A

(c) When coils, A and B are used in parallel:
 1/R = 1/R_A+ 1/R_B
Here, R_A = 24 Ω and R_B = 24 Ω
 1/R = 1/24 + 1/24
or 1/R = 2/24
      R = 12 Ω
Total resistance of parallel combination, R = 12 Ω
Now, V = IR
220 = 12 I
I = 18.33 A

Answer:

(a) When three resistors are connected in parallel, the net resistance can be obtained as followed:
1/R = 1/R₁ + 1/R₂ + 1/R₃
The resistors of 30 Ω, 20 Ω and 60 Ω are connected in parallel. Therefore, the net resistance R will be:
1/R = 1/30 + 1/20 +1/60
1/R = (2 + 3 + 1)/60
1/R = 6/60
R =10 Ω
Two more resistors of 10 Ω and 40 Ω are connected in parallel to each other. Therefore, their net resistance, R' will be:
1/R^' = (1/10) + (1/40)
1/R^' = (4 + 1)/40
1/R^' = 5/40
R' = 8 Ω
The resistors of 8 Ω and 10 Ω are connected in series. Therefore, the net resistance of the circuit =R+R^,=8 Ω + 10 Ω = 18 Ω

(b) The total current flowing through the circuit can be calculated as:
I = V/R
I = 12/18
I = 0.67 A

Answer:

I(a) Let I₁, I₂ and I₃ be the current flowing through the resistors of 5 Ω, 100 Ω and 30 Ω, respectively.
According to Ohm's law, V = IR
Here, V = 12 V and R = 5 Ω
I = V/R
I₁ = 12/5 = 2.4 A
I₂ = 12/10 = 1.2 A
I₃ = 12/30 = 0.4 A

(b) Total current through the circuit, I = I₁ + I₂ + I₃
=  2.4 A + 1.2 A + 0.4 A
= 4 A

(c) The resistors of 5 Ω, 100 Ω and 30 Ω are connected in parallel. Therefore, the net resistance will be:
 1/ R = 1/R₁ + 1/R₂ + 1/R₃
1/R = (1/5) + (1/10) + (1/30)
1/R = (6 + 3 +1)/30
I/R = 10/30
R = 3 Ω

Answer:

(a) When two resistors are connected in series, their resultant resistance is given by
R = R₁+ R₂
Here, R₁=6 $\mathrm{\Omega }$,
         R₂=2 $\mathrm{\Omega }$
Therefore
The resulatant resistance, R = 6 + 2 = 8 Ω

(b) The current flowing through circuit is:
   V = IR
   I = V/R = 4/8
   I = 0.5 A

(c) The p.d. across the 6 Ω resistor is:
   V = 0.5 A x 6 Ω = 3 V

Answer:

(a) The resistors of 3 Ω and 6 Ω are connected in parallel. Therefore, their net resistance can be calculated as:
1/R = 1/R₁ + 1/R₂
 Here, R₁ = 3 Ώ
          R₂ = 6 Ώ
So:
1/R = (1/3) + (1/6)
1/R = (2 + 1)/6
1/R = 3/6
R = 2 Ω

(b) The current flowing through the main circuit,
I = V/R
I = 6/2 A
I = 3 A

(c) The current flowing in the 3 Ω resistor,
I = V/R
I = 6/3 = 2 A

Answer:

(a) The resistors of 2 Ω and 3 Ω are connected in series.
Therefore their combined resistance, R = R₁ + R₂
Here, R₁=2 Ω
         R₂=3 Ω
So, the combined resistance, R = 2 Ω + 3 Ω
                                            R = 5 Ω

(b) The current, I in the circuit can be calculated as:
        V = IR
   or  I = 10 / 5 = 2 At)=R2R1+R2v(t)
(c) The p.d. across the 2 Ω resistor is:
      V = IR
 or V = 2$\times$2= 4 V

(d) The p.d. across the 3 Ω resistor:
     V = IR
or  V = 2 x 3
 or V = 6 V

Answer:

(a) The resistors of 6 Ω and 3 Ω are connected in parallel. Therefore, their combined resistance can be calculated as:
1/R = 1/R₁ + 1/R₂
 Here, R₁ = 6 Ω ,
          R₂ = 3 Ω
1/R = (1/6) + (1/3 )
1/R = (1 + 2)/6
R = 6/3
R = 2 Ω

(b) Potential difference, V = IR
The potential difference across the combined resistance,
V = (6 $\times$ 2) V
V = 12 V

(c) It can be seen here that the potential difference across the 3 Ω resistor is 12 V. 
 
(d) The current across the 3 Ω resistor is:
I = V/R
I = 12/3
I = 4 A

(e) The current across the 6 Ω resistor is:
I = V/R
I = 12/6
I = 2 A

Answer:

(a) Circuit diagram

(b) Since two resistors of 20 Ω  are connected in series, the effective resistance will be
R = R₁+ R₂
Here, R₁=20 Ω,
          R₂ =20 $\mathrm{\Omega }$
Therefore,
R = 20 Ω + 20 Ω = 40 $\mathrm{\Omega }$
The effective resistance of the two resistors, R= 40 Ω
(c) The current that flows from the battery,
I = V / R
I = 5 / 40
I = 0.125 A

(d) The p.d. across the 20 Ω resistor, V= I $\times$ R
V = 20 x 0.125 V
V = 2.5 V
Similarly, the potential difference across the second 20 Ω resistor is equal to 2.5 V.

Answer:

(a) The resistors are connected in parallel.
(b) In a parallel arrangement, the voltage remains the same across each resistor. The battery is also connected in parallel with the resistors. Therefore, the p.d across each resistor is 6 V.
(c) The 2 Ω resistor will have the bigger share of the current since resistance is inversely proportional to the current.
(d) The effective resistance of the parallel combination is:
$$\begin{gathered} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R}=\frac{1}{2}+\frac{1}{3} \\ \frac{1}{R}=\frac{3+2}{6} \\ \frac{1}{R}=\frac{5}{6} \\ R=\frac{6}{5}\mathrm{\Omega } \\ R=1.2 \mathrm{\Omega } \end{gathered}$$

(e) The current flowing through the battery, I = 5 A

Answer:

The coils of the resistors of 4 Ω and 2 Ω are connected in parallel. Therefore:
$$\begin{gathered} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R}=\frac{1}{4}+\frac{1}{2} \\ \frac{1}{R}=\frac{3}{4} \\ R=\frac{4}{3}\mathrm{\Omega } \end{gathered}$$
Combined resistance of the 4 Ω coil and the 2 Ω coil = 1.33 Ω
Total current = 3 A
The voltage across the whole cicuit,as in case of parallel circuit volatge is same scross all resistances.
V = IR.
V = 3 x 4/3 = 4 V
Therefore, the current through the 2 Ω coil,
I = V/R₁.
I = 4/2 A
I = 2 A

Answer:

(a)
 
(a) Let the current in the circuit be I amperes and the battery be of strength V volts.
Let the combined resistance of the three resistors be R ohms.
Therefore, according to Ohm's law, we have:
V = IR ... (1)
We know that when resistors are connected in series, the current is the same in all the resistors but the voltage is different across each resistor.
Therefore:
V = V₁ + V₂
V = IR₁ + IR₂
V = I (R₁ + R₂) ... (2)
From the equations (1) and (2) we have:
R = R₁ + R₂

(b)
(i) The current through the 5 Ω resistance can be obtained using the equation,
I = V/R
or I = 10/5 = 2 A
(ii) In a series arrangement, the current remains the same across each resistance. Therefore, the current through R also will be 2 A.

(iii) Value of R = V/I
                    R = 6/2 = 3 Ω
(iv) The value of V = IR
V = 2 x (3 + 5)  = 16 V

Answer:

(a)

Let the current in the circuit be I amperes and the battery be of strength V volts.
Let the combined resistance of the three resistors be R ohms.
Therefore, according to Ohm's law, we have:
V = IR .... (1)
We know that when the resistors are connected in series, the current is the same in all the resistors. Therefore:
V = V₁ + V₂ + V₃
V = IR₁ + IR₂ + IR₃
V = I (R₁ + R₂ + R₃) ... (2)
From the equations (1) and (2) we have:
R = R₁ + R₂ + R₃

(b) The current through the 5 Ω resistor can be calculated as:
V = IR
I = V/R
I = 6/5 = 1.2 A
The current through the 10 Ω resistor can be calculated as:
I = V/R
I = 6/10 = 0.6 A
The current through the 30 Ω resistor can be calculated as:
I = V/R
I = 6/30 = 0.2 A
(b)The total current in the circuit, I = 1.2 A + 0.6 A + 0.2 A = 2 A

(c) The three resistors are connected in parallel. Therefore:
1/R = 1/R₁ + 1/R₂ + 1/R₃
Here, R₁=5 $\mathrm{\Omega }$,
         R₂ = 10 $\mathrm{\Omega }$,
         R₃= 30 $\mathrm{\Omega }$
1/R = (1/5) + (1/10) + (1/30)
1/R = (6 + 3 + 1)/30
1/R = 10/30
R = 3 Ω

Answer:

(a)

(a) Let the individual resistance of the two resistors be R₁ and R₂ and their combined resistance be R. Let the total current flowing in the circuit be I and strength of the battery be V volts. Then, from Ohm's law, we have:
V = IR ... (1)
We know that when resistors are connected in parallel, the potential drop across each resistance is the same.
Therefore:
I  = I₁ + I₂
I = V/R₁ + V/R₂
I = V/(1/R₁ + 1/R₂) ... (2)
From the equations (1) and (2) we have:
1/R = 1/R₁ + 1/R₂

(b)
(1)
 $$\begin{gathered} \mathrm{As} \mathrm{shown} \mathrm{Resistance} 2 \mathrm{\Omega } \mathrm{and} 3 \mathrm{\Omega } \mathrm{are} \mathrm{in} \mathrm{series} \\ \mathrm{we} \mathrm{know} \mathrm{resistance} \mathrm{in} \mathrm{series} \mathrm{arrangement} \mathrm{can} \mathrm{be} \mathrm{obtained} \mathrm{as} R=R_1+R_2 \\ R=2 \mathrm{\Omega }+3 \mathrm{\Omega }=5 \mathrm{\Omega } \\ \mathrm{Since} \mathrm{this} 5 \mathrm{\Omega } \mathrm{is} \mathrm{in} \mathrm{parallel} \mathrm{with} \mathrm{another} 5 \mathrm{\Omega } \mathrm{resistance} \mathrm{therefore} \mathrm{total} \mathrm{resistance} \mathrm{can} \mathrm{be} \mathrm{obtained} \mathrm{as} \\ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \mathrm{Here} {\mathrm{R}}_1=5 \mathrm{\Omega }, {\mathrm{R}}_2=5 \mathrm{\Omega } \\ \frac{1}{R}=\frac{1}{5}+\frac{1}{5} \\ \frac{1}{R}=\frac{2}{5} \\ R =2.5 \mathrm{\Omega } \\ \mathrm{Total} \mathrm{resistance} \mathrm{of} \mathrm{combination}=2.5 \mathrm{\Omega } \end{gathered}$$

(ii) The current shown by the ammeter A, i.e. the current in the circuit can be calculated as:
      I = $\frac{\mathrm{V}}{\mathrm{R}}$

     I =$\frac{4}{2.5}$= 1.6 A

Answer:

 (d) 7$\frac{1}{2}$$\mathrm{\Omega }$
The resistors of 6 Ώ and 2 Ώ are connected in parallel.
$$\begin{gathered} \therefore \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \mathrm{Here}, \\ R_{1 }= 6 \mathrm{\Omega } \\ R_2=2 \mathrm{\Omega } \\ \frac{1}{R}=\frac{1}{6}+\frac{1}{2} \\ \frac{1}{R}=\frac{4}{6} \\ R=\frac{6}{4} \end{gathered}$$

This arrangement is further connected in series with the 6 Ώ resistor.
∴ Net resistance =$\frac{6}{4}+6$ =$7\frac{1}{2}$ Ω

Answer:

 (c) 20 Ω
When 25 Ω and 15 Ω are connected in series, then:
Total resistance, R = 25 + 15 = 40 Ώ
This 40 Ώ is connected in parallel with the 40 Ώ resistor.
Therefore, the net resistance = 40 / 2 = 20 Ώ

Answer:

 (a) 9 Ώ
The two resistors of 6 Ώ are connected in parallel with each other. So, their net resistance 3 Ώ is connected in series with a resistance of 6 Ώ. So, the net resistance of the complete arrangement is 9 Ω.

Answer:

 (c) 1V
The resistors of 1 $\mathrm{\Omega }$, 2 $\mathrm{\Omega }$ and 3 $\mathrm{\Omega }$  are connected in series. Therefore, the net resistance,
R =R₁ +R₂ +R₃
R=1 $\mathrm{\Omega }$ + 2 $\mathrm{\Omega }$ + 3 $\mathrm{\Omega }$ = 6 $\mathrm{\Omega }$
Current in the circuit will be,
I = V/R
or I = 2/6 = 1/3 A
Current = $\frac{1}{3}$A
Therefore, the voltage across the 3 Ώ resistor,
V = IR
or V= $\frac{1}{3}$ x 3 = 1 V