Hc Verma II for Class 11 Science Physics Chapter 39 - Alternating Current

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Page No 328:

Answer:

The reactance of a capacitor is given by,
$X_{c} = \frac{1}{ω C}$
For a DC source, ω = 0
$\Rightarrow X_{c} = \frac{1}{0 \times C} = \infty$
So, for a constant DC source, reactance of a capacitor is infinite.

Page No 328:

Answer:

Voltage, V = V₀cos ωt
Current, i = i₀ sin ωt or i = i₀ cos (ωt $-$ $\frac{π}{2}$ )

Power dissipated in an AC circuit is given by,
$P = I_{r m s} V_{r m s} \cos ϕ$ ,
where I_rms = rms value of current
V_rms= rms value of voltage
ϕ = phase difference between current and voltage
Here, ϕ = $π$ /2
$\Rightarrow \cos ϕ = \cos \frac{π}{2} = 0 ∴ P = I_{r m s} V_{r m s} \times 0 = 0$

Page No 328:

Answer:

The rms value of current is given by,
$i_{r m s} = \frac{i_{0}}{\sqrt{2}}$
Since peak value of current i₀ is same for both currents, their rms values will be same.

Page No 328:

Answer:

Let a LCR circuit is connected across an AC supply with the emf E = E₀ sin ωt.
Let the inductance in the circuit be L
Let the net impedence of the circuit be $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$
Where,
R = resistance in the circuit
X_L= reactance due to inductor
X_C= reactance due to capacitor
The magnitude of the voltage across the inductor is given by
$V = L \frac{d i}{d t}$
The current in the circuit can be written as $I = I_{0} \sin (ω t + ϕ)$
Where, ϕ is the phase difference between the current and the supply voltage
Thus, the voltage across the inductor can be written as
$V = L I_{0} \cos (ω t + ϕ)$
Thus peak value of the voltage across the inductor is given by
$V = L I_{0} \Rightarrow V = \frac{E_{0}}{Z} \times L$
Therefore, the peak voltage across the inductor is given by $V = \frac{E_{0}}{Z} \times L$
At resonance Z = R,
$V = \frac{E_{0}}{R} \times L$ ,
If $\frac{L}{R} > 1$
V > E₀
Therefore if magnitude of $\frac{L}{R} > 1$ at resonance the value of the voltage across the inductor will bw greater than the peak value of the supply voltage.

Page No 328:

Answer:

Ohm's Law is valid for resistive circuits only. It is not valid for capacitive or inductive circuits, or a combination of both.

Page No 328:

Answer:

The reactance of a capacitor is given by,
$X_{c} = \frac{1}{ω C}$
Also,
$C = \frac{K ε_{0} A}{d}$ ,
where C = capacitance
K = dielectric constant
A = area of plates
d = distance between the plates.
$K > 1$
$∴$ The capacitance C of the capacitor will increase on inserting the dielectric slab and, consequently, the reactance X_c will decrease.
Rms current, $i_{r m s} = \frac{ε_{0}}{\sqrt{2} X_{C}}$

Therefore, rms current will decrease.

Page No 328:

Answer:

The condition for resonance is:
$\frac{1}{ω C} = ω L$
The peak current through the circuit is given by,
$i_{0} = \frac{V_{0}}{{\sqrt{R^{2} + (\frac{1}{ω C} - ω L)}}^{2}}$
From the condition of resonance, we get:
$i_{0} = \frac{V_{0}}{R}$
The current will flow through the all circuit elements. But since the reactance of the capacitor and inductor are equal, the potential difference across them will be equal and opposite and will cancel each other.

Page No 328:

Answer:

No. When an AC source is connected to a capacitor, there is a steady in the circuit to transfer change to the plates of the capacitor. This produces a potential difference between the plates. The capacitance is alternatively charged and discharged as the current reverses after each half cycle.

Page No 329:

Answer:

The thermal energy produced for an AC circuit in one time period is given by,
$H = {I_{r m s}}^{2} \times R \times \frac{2 π}{ω}$
For current, i₁ = i₀ sin ωt,
$I_{r m s} = \frac{i_{0}}{\sqrt{2}}$
$\Rightarrow H = \frac{{i_{0}}^{2} R}{2} \times \frac{2 π}{ω} = \frac{π {i_{0}}^{2} R}{ω}$

For current, i₂ = −i₀ sin ωt,
$I_{r m s} = \frac{i_{0}}{\sqrt{2}}$
Hence, the same thermal energy will be produced due to this current.

Since, the direction of i₁ and i₂ are opposite and their magnitude is same, the net current through the resistor will become zero when both are passed together. Yes, the principle of superposition is obeyed in this case.

Page No 329:

Answer:

When a transformer steps up the voltage, the voltage increases but current decreases. Neglecting any loss of energy, the power remains constant and, hence, energy is not produced. It remains constant.

Page No 329:

Answer:

A transformer is ideally an inductive coil. For an inductor connected across a DC voltage,

$V - L \frac{d i}{d t} = 0 \Rightarrow V = L \frac{d i}{d t} \Rightarrow \int d i = \frac{V}{L} \int d t \Rightarrow i = \frac{V t}{L}$

For a DC source, the current across the inductor will increase with time and can reach a very large value, which can burn the transformer.

Page No 329:

Answer:

Let us consider an AC series LCR circuit of angular frequency ω. The impedance of the circuit is given by,
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
The phase difference between V and I is given by,
$\tan ϕ = (\frac{ω L - \frac{1}{ω C}}{Z})$
From the above formula, we can clearly see that
$ϕ \in (- \frac{π}{2}, \frac{π}{2})$
So, we cannot have a phase difference of 180° or 120°.

Page No 329:

Answer:

When a capacitor is included in a series circuit, the impedance of the circuit,
$Z = \sqrt{R^{2} + {X_{C}}^{2}}$
The power absorbed by a resistor is given by,
$P = {I_{r m s}}^{2} R$
Since impedance increases due to introduction of a capacitor, the rms value of current I_rms will decrease and, hence, the power absorbed by the resistor will decrease.

When a small inductance is introduced in the circuit, the impedance of the circuit,
$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}$
Since the impedance now decreases a little, the rms value of current will increase and, hence, the power absorbed by the resistor will increase.

Page No 329:

Answer:

A hot-wire ammeter measures the rms value of current for an alternating current. So, it can be used to measure the direct current of constant value because that constant value will be equal to the rms value of current. As, the rms value of the current is same as the direct current thus we need not change the graduations.

Page No 329:

Answer:

(a) DC

For a DC source, ω = 0. So, the reactance of the capacitance is given by,
$X_{C} = \frac{1}{ω C} = \frac{1}{0} = \infty$

Page No 329:

Answer:

(c) i₁ < i₂

The charge on the capacitor during steady state is given by,
$Q = C ε = ε_{0} C [\cos (100 {πs}^{- 1}) t + \cos (500 {πs}^{- 1}) t]$

The steady state current is, thus, given by,
$i = \frac{d Q}{d t} = ε_{0} C \times 100 π [\sin (100 {πs}^{- 1}) t] + ε_{0} C \times 500 π [\sin (500 {πs}^{- 1}) t] \Rightarrow i = 100 C {πε}_{0} \cos [(100 {πs}^{- 1}) t + ϕ_{1}] + 500 C {πε}_{0} \cos [(500 {πs}^{- 1}) + ϕ_{2}] \Rightarrow i_{1} = 100 C {πε}_{0} & i_{2} = 500 C {πε}_{0} ∴ i_{2} > i_{1}$

Page No 329:

Answer:

(c) about 310 V

Given:
V_rms = 220 V
The peak value of voltage is given by,
$V_{p} = \sqrt{2} \times V_{r m s} = 1.414 \times 220 = 311 V$

Page No 329:

Answer:

(b) may be zero

Let the AC voltage be given by,
$V = V_{0} \sin ω t$
Here, ω = 2 $π$ f = 314 rad/s
The average voltage over the given time,
$V_{a v g} = \frac{\int_{0}^{0.01} V d t}{\int_{0}^{0.01} d t} = - V_{0} {[\frac{\cos ω t}{ω}]}_{0}^{0.01} = \frac{V_{0}}{ω \times 0.01} (1 - \cos ω (0.01)) = \frac{V_{0}}{314 \times 0.01} (1 - \cos (314 \times 0.01)) = \frac{V_{0}}{3.14} (1 - cosπ) = \frac{2 V_{0}}{π} = 140.127 V$

Also, when $V = V_{0} \cos ω t$ ,
$V_{a v g} = \frac{\int_{0}^{0.01} V d t}{\int_{0}^{0.01} d t} = V_{0} {[\frac{\sin ω t}{ω}]}_{0}^{0.01} = \frac{V_{0}}{ω \times 0.01} (\sin ω (0.01) - 0) = \frac{V_{0}}{314 \times 0.01} (\sin (314 \times 0.01)) = \frac{V_{0}}{3.14} (\sin π) = 0$

From the above results, we can say that the average voltage can be zero. But it is not necessary that it should be zero or never zero.

Page No 329:

Answer:

(b) 50 Hz

The magnetic field energy in an inductor is given by,
$E = \frac{1}{2} L i^{2}$
The magnetic energy will be maximum when the current will reach its peak value, i₀, and it will be minimum when the current will become zero.

From the above graph of alternating current, we can see that current reduces from maximum to zero in T/4 time, where T is the time period.
So, in this case, T/4 = 5 ms
$\Rightarrow T = 20 ms ∴ Frequency, ν = \frac{1}{T} = \frac{1}{20 \times 10^{- 3}} = 50 Hz$

Page No 329:

Answer:

(d)

The reactance of a series LC circuit is given by,
$X = X_{L} - X_{C} = ω L - \frac{1}{ω C} \Rightarrow X = 2 π f L - \frac{1}{2 π f C}$
The correct relation between the reactance and f is represented by the graph in option (d).
$X = 0$ , when $X_{L} = X_{C}$ .

Thus plot d represent the reactance of a series LC combination.

Page No 329:

Answer:

(a) 5 Ω

The impedance of the circuit is given by
$Z = \sqrt{R^{2} + X^{2}} R = 4 Ω & X = 3 Ω \Rightarrow Z = \sqrt{4^{2} + 3^{2}} = 5 Ω$

Page No 329:

Answer:

(b) in AC circuits only

When a DC supply is provided to a transformer, there will be no change in flux with time across the coils of the transformer. So, there will be no induced emf in the secondary coil due to changing current in the primary coil. Hence, the transformer cannot operate in DC because of the violation of its working principle.

Page No 329:

Answer:

(c) $\sqrt{\frac{i_{1}^{2} + i_{2}^{2}}{2}}$
Given:
i = i₁ cos ωt + i₂ sin ωt
The rms value of current is given by,
$i_{r m s} = \sqrt{\frac{\int_{0}^{T} i^{2} d t}{\int_{0}^{T} d t}} i = i_{1} \cos ω t + i_{2} \sin ω t i_{rms} = \sqrt{\frac{\int_{0}^{T} {(i_{1} \cos ω t + i_{2} \sin ω t)}^{2} d t}{\int_{0}^{T} d t}} i_{rms} = \sqrt{\frac{\int_{0}^{T} (i_{1}^{2} \cos^{2} ω t + i_{2}^{2} \sin^{2} ω t + 2 i_{1} i_{2} \sin ωt \cos ω t) d t}{\int_{0}^{T} d t}} i_{rms} = \sqrt{\frac{\int_{0}^{T} (i_{1}^{2} \frac{(\cos 2 ω t + 1)}{2} + i_{2}^{2} \frac{(1 - \cos 2 ω t)}{2} + i_{1} i_{2} \sin 2 ω t) d t}{\int_{0}^{T} d t}} [∵ \cos^{2} ω t = \frac{(\cos 2 ω t + 1)}{2}, \sin^{2} ω t = \frac{(1 - \cos 2 ω t)}{2}]$

We know that, T = 2π

Integrating the above expression
$i_{rms} = \sqrt{\frac{\frac{1}{2} i_{1}^{2} (\int_{0}^{2 π} 1 d t + \int_{0}^{2 π} \cos 2 ω t d t) + i_{2}^{2} (\int_{0}^{2 π} 1 d t - \int_{0}^{2 π} \cos 2 ω t d t) + i_{1} i_{2} \int_{0}^{2 π} \sin 2 ω t d t}{\int_{0}^{2 π} d t}}$

The following integrals become zero
$\int_{0}^{2 π} \cos 2 ω t d t = 0 = \int_{0}^{2 π} \sin 2 ω t d t$
Therefore, it becomes
$i_{rms} = \sqrt{\frac{\frac{i_{1}^{2}}{2} (\int_{0}^{2 π} 1 d t) + \frac{i_{2}^{2}}{2} (\int_{0}^{2 π} 1 d t)}{\int_{0}^{2 π} d t}} i_{rms} = \sqrt{\frac{\frac{i_{1}^{2}}{2} \times 2 π + \frac{i_{2}^{2}}{2} \times 2 π}{2 π}} \Rightarrow i_{rms} = \sqrt{\frac{i_{1}^{2} + i_{2}^{2}}{2}}$

Page No 329:

Answer:

(d) about 10 A

The rms value of an alternating current is equivalent to the constant current. So, the heating effect produced is actually measured in terms of the rms value, in case of alternating current. The constant current is, thus, equal to the rms value of alternating current, which is given by,
$I_{r m s} = \frac{I_{p e a k}}{\sqrt{2}} = \frac{14}{\sqrt{2}} = 9.9 \approx 10 A$

Page No 329:

Answer:

(a) 2.8 A

The constant current is equal to the rms value of current. So,
I_rms= 2.8 A

Page No 329:

Answer:

(a) of the inductor increases

The reactance of an inductor is given by,
$X_{L} = ω L$
And the reactance of a capacitor is given by,
$X_{C} = \frac{1}{ω C}$
Here, ω = 2πf , where f is the frequency of the source. So, when f increases, ω increases.
∴ X_L will increase and X_C will decrease.

Page No 329:

Answer:

(a) an inductor and a capacitor
(d) neither an inductor nor a capacitor

(a) The reactance of a circuit containing a capacitor and an inductor is given by,
$X = X_{L} - X_{C} = ω L - \frac{1}{ω C}$
If $ω L = \frac{1}{ω C}$ , X = 0. So, the circuit contains an inductor and a capacitor.

(b) For a circuit without a capacitor, reactance is given by,
$X^{'} = X_{L} = ω L$ , which cannot be zero for an AC source.

(c) Similarly, for a circuit without an inductor, reactance is given by,
$X^{''} = X_{C} = \frac{1}{ω C}$ , which also cannot be zero.

(d) For a circuit without any capacitor and inductor, reactance, X^' = 0 ( L = C = 0)

Page No 330:

Answer:

(a) pure inductor
(b) pure capacitor
(d) combination of an inductor and a capacitor.

For a pure inductive circuit, voltage leads the current by $90 °$ . So, the instantaneous current is zero when the instantaneous voltage is maximum.
Similar is the case with a purely capacitive circuit, in which, current leads the voltage by $90 °$ .

Also, in a circuit containing a combination of inductor and capacitor, the current may lead or lag the voltage by $90 °$ , depending upon whether the voltage across the inductor or the capacitor is greater. Here too, there is a phase difference of $90 °$ between the voltage and current. Hence, the the instantaneous current is zero when the instantaneous voltage is maximum.

Page No 330:

Answer:

(a) Current
(b) Induced emf in the inductor

For a series L-R circuit, the AC current can be given by,
$i = i_{0} \sin ω t$

From the graph, we can see that the average value of current over a cycle is zero.

Since it a series L-R circuit, the phase difference between current and voltage is $\frac{π}{2}$ . The AC voltage can be given by,
$V = V_{0} \cos ω t$

From the graph, we can see that the average value of voltage over a cycle is also zero.

Joule's heat through the resistor is given by,
$H_{a v g} = {i_{r m s}}^{2} R$ , which is non zero.

Similarly, magnetic energy stored in the inductor is given by,
$U_{a v g} = \frac{1}{2} L {i_{r m s}}^{2}$ , which is also non-zero.

Page No 330:

Answer:

(b) a hot-wire voltmeter

Only a hot-wire voltmeter can be used to measure an AC voltage across a resistor.

Page No 330:

Answer:

(a) DC dynamo
(b) AC dynamo

An AC or DC dynamo can be used to convert mechanical energy to electrical energy.
A motor converts electrical energy to mechanical energy and a transformer is used to step up or down the voltage or simply to transfer electrical energy.

Page No 330:

Answer:

(b) may be 1000 W
(d) may be less than 1000 W

The average power delivered by an AC source is given by,
$P_{a v g} = V_{r m s} I_{r m s} \cos ϕ$
Given:
V_rms = 100 V
I_rms = 10 A
$\Rightarrow P_{a v g} = 1000 \cos ϕ$
$∵ ϕ \in (- \frac{π}{2}, \frac{π}{2})$
$\Rightarrow \cos ϕ \in (0, 1)$
$∴ 0 \leq P_{a v g} \leq 1000$

Page No 330:

Answer:

Frequency of alternating current, f = 50 Hz
Alternation current $(i)$ is given by,
i = i₀sinωt ...(1)
Here, i₀ = peak value of current
Root mean square value of current $(i_{r m s})$ is given by,
$i_{rms} = \frac{i_{0}}{\sqrt{2}} . . . (1)$
On substituting the value of the root mean square value of current in place of alternating current in equation (1), we get:
$\frac{i_{0}}{\sqrt{2}} = i_{0} \sin ω t \Rightarrow \frac{1}{\sqrt{2}} = \sin ω t = \sin \frac{π}{4} \Rightarrow \frac{π}{4} = ω t \Rightarrow t = \frac{π}{4 ω} = \frac{π}{4 \times 2 π f} (∵ ω = 2 π f) = \frac{1}{8 f} = \frac{1}{8 \times 50} = \frac{1}{400} = 0.0025 s = 2.5 ms$

Page No 330:

Answer:

RMS value of voltage, E_rms = 220 V,
Frequency of alternating current, f = 50 Hz
(a) Peak value of voltage $(E_{0})$ is given by,
$E_{0} = E_{rms} \sqrt{2}$ ,
where E_rms= root mean square value of voltage
$E_{0} = E_{rms} \sqrt{2} \Rightarrow E_{0} = \sqrt{2} \times 220 \Rightarrow E_{0} = 311.08 V = 311 V$
(b) Voltage $(E)$ is given by,
$E = E_{0} \sin ω t$ ,
where E₀ = peak value of voltage
Time taken for the current to reach zero from the rms value = Time taken for the current to reach the rms value from zero
In one complete cycle, current starts from zero and again reaches zero.
So, first we need to find the time taken for the current to reach the rms value from zero.
$As E = \frac{E_{0}}{\sqrt{2}}, \frac{E_{0}}{\sqrt{2}} = E_{0} \sin ω t \Rightarrow ω t = \frac{π}{4} \Rightarrow t = \frac{π}{4 ω} = \frac{π}{4 \times 2 π f} \Rightarrow t = \frac{π}{8 π 50} = \frac{1}{400} \Rightarrow t = 2.5 ms$
Thus, the least possible time in which voltage can change from the rms value to zero is 2.5 ms.

Page No 330:

Answer:

Power of the bulb, P = 60 W
Voltage at the bulb, V = 220 V
RMS value of alternating voltage, E_rms = 220 V
P = V²R,
where R = resistance of the bulb
$∴ R = \frac{V^{2}}{P} = \frac{220 \times 220}{60} = 806.67$
Peak value of voltage $(E_{0})$ is given by,
$E_{0} = E_{rms} \sqrt{2}$
= $220 \times \sqrt{2}$
= 311.08
Now, maximum current through the filament $(i_{0})$ is,
$i_{0} = \frac{E_{0}}{R}$
$\Rightarrow i_{0} = \frac{311.08}{806.67} = 0.39 A$

Page No 330:

Answer:

Voltage across the electric bulb, E = 12 volts
Let E₀ be the peak value of voltage.
We know that heat produced by passing an alternating current $(i)$ through a resistor is equal to heat produced by passing a constant current $(i_{r m s})$ through the same resistor. If R is the resistance of the electric bulb and T is the temperature, then
$i^{2} R T = i_{rms}^{2} R T \Rightarrow \frac{E^{2}}{R^{2}} = \frac{E_{rms}^{2}}{R^{2}} \Rightarrow E^{2} = \frac{E_{0}^{2}}{2} (∵ {E^{2}}_{r m s} = \frac{{E^{2}}_{0}}{2}) \Rightarrow E_{0}^{2} = 2 E^{2} \Rightarrow E_{0}^{2} = 2 \times {(12)}^{2} = 2 \times 144 \Rightarrow E_{0} = \sqrt{2 \times 144} = 16.97 = 17 V$
Thus , peak value of voltage is 17 V.

Page No 330:

Answer:

Peak power of the resistive coil, $P_{0} = 80 W$
Time, t = 100 s
RMS value of power $(P_{r m s})$ is given by,
$P_{rms} = \frac{P_{0}}{2}$ ,
where P₀ = Peak value of power
$∴ P_{rms} = \frac{P_{0}}{2} = 40 W$
Energy consumed $(E)$ is given by,
E = P_rms × t
= 40 × 100
= 4000 J = 4.0 kJ

Page No 330:

Answer:

Given:
Area of parallel-plate air-capacitor, A = 20 cm²
Separation between the plates, d = 0.1 mm
Dielectric strength of air, E= 3 × 10⁶ V/m
E = $\frac{V}{d}$ ,
where V = potential difference across the capacitor
$∴$ V = Ed
= 3 × 10⁶× 0.1 × 10⁻³
= 3 × 10² = 300 V
Thus, peak value of voltage is 300 V.
Maximum rms value of voltage $(V_{rms})$ is given by,
$V_{rms} = \frac{V_{0}}{\sqrt{2}} = \frac{300}{\sqrt{2}} = 212 V$

Page No 330:

Answer:

As per the question,
$i = i_{0} e^{\frac{- t}{τ}}$
We need to find the rms current. So, taking the average of i within the limits 0 to τ and then dividing by the given time period τ, we get:
${i_{rms}}^{2} = \frac{1}{τ} \int_{0}^{τ} i_{0}^{2} e^{- 2 t / τ} d t \Rightarrow {i_{r m s}}^{2} = \frac{i_{0}^{2}}{τ} \int_{0}^{τ} e^{- 2 t / τ} d t = \frac{i_{0}^{2}}{τ} \times {[\frac{- τ}{2} e^{- 2 t / τ}]}_{0}^{τ} = - \frac{i_{0}^{2}}{τ} \times \frac{τ}{2} \times [e^{- 2} - 1] = \frac{{i_{0}}^{2}}{2} (1 - \frac{1}{e^{2}}) i_{rms} = \frac{i_{0}}{e} \sqrt{\frac{e^{2} - 1}{2}}$

Page No 330:

Answer:

Capacitance of the capacitor, C = 10 μF = 10 × 10⁻⁶ F = 10⁻⁵ F
Output voltage of the oscillator, $ε$ = (10 V)sinωt
On comparing the output voltage of the oscillator with $ε = ε_{0} \sin ω t$ , we get:
Peak voltage $ε_{0}$ = 10 V
For a capacitive circuit,
Reactance, $X_{c} = \frac{1}{ω C}$
Here, $ω$ = angular frequency
C = capacitor of capacitance
Peak current, $I_{0}$ = $\frac{ε_{0}}{X_{c}}$
(a) At ω = 10 s⁻¹:
Peak current,
I₀ = $\frac{ε_{0}}{X_{c}}$
$= \frac{ε_{0}}{1 / ω C} = \frac{10}{1 / 10 \times 10^{- 5}} A$
= 1 × 10⁻³ A
(b) At ω = 100 s⁻¹:
Peak current, I₀ = $\frac{ε_{0}}{1 / ω C}$
$\Rightarrow I_{0} = \frac{10}{1 / 100 \times 10^{- 5}} \Rightarrow I_{0} = \frac{10}{10^{3}} = 1 \times 10^{- 2} A = 0.01 A$

(c) At ω = 500 s⁻¹:
Peak current, I₀ = $\frac{ε_{0}}{\frac{1}{ω C}}$
$I_{0} = \frac{ε_{0}}{1 / ω C} \Rightarrow I_{0} = \frac{10}{1 / 500 \times 10^{- 5}} \Rightarrow I_{0} = 10 \times 500 \times 10^{- 5} = 5 \times 10^{- 2} A = 0.05 A$

(d) At ω = 1000 s⁻¹:
Peak current, I₀ = $\frac{ε_{0}}{\frac{1}{ω C}}$
$\Rightarrow I_{0} = \frac{10}{1 / 1000 \times 10^{- 5}} \Rightarrow I_{0} = 10 \times 1000 \times 10^{- 5} \Rightarrow I_{0} = 10^{- 1} A = 0.1 A$

Page No 330:

Answer:

Given:
Inductance of the coil, L = 5.0 mH = 0.005 H
(a) At ω = 100 s⁻¹:
Reactance of coil $(X_{L})$ is given by,
X_L = $ω L$
Here, $ω$ = angular frequency
$∴$ X_L = 100 $\times$ 0.005 = 0.5 $Ω$
Peak current, I₀ = $\frac{10}{0.5} = 20 A$
(b) At ω = 500 s⁻¹:
$Reactance, X_{L} = 500 \times \frac{5}{1000} = 2.5 Ω$
Peak current, I₀ = $\frac{10}{2.5} = 4 A$
(c) ω = 1000 s⁻¹:
Reactance, X_L= 1000 $\times 0.005$ = 5 $Ω$
Peak current, I₀ = $\frac{10}{5}$ = 2 A

Page No 330:

Answer:

Given:
Resistance of coil, R = 10 Ω
Inductance of coil, L = 0.4 Henry
Voltage of AC source, Erms = 6.5 V
Frequency of AC source, $f$ = $\frac{30}{π} Hz$
Reactance of resistance-inductance circuit $(Z)$ is given by,
Z = $\sqrt{R^{2} + {X_{L}}^{2}}$
Here, R = resistance of the circuit
X_L = Reactance of the pure inductive circuit
Z = $\sqrt{R^{2} + {(2 π f L)}^{2}}$
= $\sqrt{{(10)}^{2} + {(2 \times π \times \frac{30}{π} \times 0.4)}^{2}}$
Average power consumed in the circuit (P) is given by,
P = E_rmsI_rmscos $ϕ$
cos $ϕ$ = $\frac{R}{Z}$ , I_rms = $\frac{E_{r m s}}{Z}$
$∴ P = 6.5 \times \frac{6.5}{Z} \times \frac{R}{Z} \Rightarrow P = \frac{6.5 \times 6.5 \times 10}{{(\sqrt{R^{2} + {(ω L)}^{2}})}^{2}} \Rightarrow P = \frac{6.5 \times 6.5 \times 10}{100 + 576} \Rightarrow P = \frac{6.5 \times 6.5 \times 10}{676} \Rightarrow P = 0.625 = \frac{5}{8} W$

Page No 330:

Answer:

Given:
Peak voltage of AC source, E₀ = 12 V
Angular frequency, ω = 250 $π$ s⁻¹
Resistance of resistor, R = 100 Ω
Energy dissipated as heat $(H)$ is given by,
$H = \frac{{E_{rms}}^{2}}{R} T$
Here, E_rms = RMS value of voltage
R = Resistance of the resistor
T = Temperature
Energy dissipated as heat during t = 0 to t = 1.0 ms,
$H = \int_{0}^{t} d H = \int \frac{E_{0}^{2} \sin^{2} ω t}{R} d t (∵ E_{r m s} = E_{0} \sin ω t) = \frac{144}{100} \int_{0}^{10^{- 3}} \sin^{2} ω t d t = 1.44 \int_{0}^{10^{- 3}} (\frac{1 - \cos 2 ω t}{2}) d t = \frac{1.44}{2} [\int_{0}^{10^{- 3}} d t + \int_{0}^{10^{- 3}} \cos 2 ω t d t] = 0.72 [10^{- 3} - {\{\frac{\sin 2 ω t}{2 ω}\}}_{0}^{10^{- 3}}] = 0.72 [\frac{1}{1000} - \frac{1}{500 π}] = 0.72 [\frac{1}{1000} - \frac{2}{1000 π}] = (\frac{π - 2}{1000 π}) \times 0.72 = 0.0002614 = 2.61 \times 10^{- 4} J$

Page No 330:

Answer:

Given:
Resistance of the series RC circuit, R = 300 Ω
Capacitance of the series RC circuit, C = 25 μF
Peak value of voltage, ε₀ = 50 V
Frequency of the AC source, ν = 50/ $π$ Hz
Capacitive reactance $(X_{C})$ is given by,
$X_{C} = \frac{1}{ω C}$
Here, $ω$ = angular frequency of AC source
            C = capacitive reactance of capacitance
$∴ X_{C} = \frac{1}{2 π \times \frac{50}{π} \times 25 \times 10^{- 6}} \Rightarrow X_{C} = \frac{10^{4}}{25} Ω$
Net reactance of the series RC circuit $(Z)$ = $\sqrt{R^{2} + {(X_{C})}^{2}}$
$\Rightarrow$ Z = $\sqrt{{(300)}^{2} + {(\frac{10^{4}}{25})}^{2}}$
      = $\sqrt{{(300)}^{2} + {(400)}^{2}}$ = 500 $Ω$
(a) Peak value of current $(I_{0})$ is given by,
$I_{0} = \frac{ε_{0}}{Z} \Rightarrow I_{0} = \frac{50}{500} = 0.1 A$

(b) Average power dissipated in the circuit $(P)$ is given by,
      P = $ε_{rms}$ I_rms cosϕ.
      $ε_{r m s}$ = $\frac{ε_{0}}{\sqrt{2}}$
       and $I_{rms} = \frac{I_{0}}{\sqrt{2}}$
$∴ P = \frac{E_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \times \frac{R}{Z} \Rightarrow P = \frac{50 \times 0.1 \times 300}{2 \times 500} \Rightarrow P = \frac{3}{2} = 1.5 W$

Page No 330:

Answer:

Power consumed by the electric bulb, P = 55 W
Voltage at which the bulb is operated, V= 110 V
Voltage of the line, V = 220 V
Frequency of the source, v = 50 Hz
P = $\frac{V^{2}}{R}$ ,
where R = resistance of electric bulb
$∴ R = \frac{V^{2}}{P}$
$= \frac{110 \times 110}{55} = 220 Ω$
If L is the inductance of the coil, then total reactance of the circuit (Z) is given by,
Z = $\sqrt{R^{2} + {(ω L)}^{2}}$
= $\sqrt{{(220)}^{2} + {(100 π L)}^{2}}$ $(∵ ω = 2 π f)$
Here, $ω$ = angular frequency of the circuit
Now, current through the bulb, $I = \frac{V}{Z}$
$∴$ Voltage drop across the bulb, V = $\frac{V}{Z} \times R$
As per question,
$110 = \frac{220 \times 220}{\sqrt{{(220)}^{2} + {(100 π L)}^{2}}} 110 = \frac{{(220)}^{2}}{\sqrt{{(220)}^{2} + {(100 π L)}^{2}}} \Rightarrow 220 \times 2 = \sqrt{{(220)}^{2} + {(100 π L)}^{2}} \Rightarrow {(220)}^{2} + {(100 π L)}^{2} = {(440)}^{2} \Rightarrow 48400 + 10^{4} π^{2} L^{2} = 193600 \Rightarrow 10^{4} π^{2} L^{2} = 193600 - 48400 \Rightarrow L^{2} = \frac{145200}{π^{2} \times 10^{4}} = 1.4726 \Rightarrow L = 1.2135 ≅ 1.2 H$

Page No 330:

Answer:

Given:
Resistance in series LCR circuit, R = 300 Ω
Capacitance in series LCR circuit, C = 20 μF= 20 × 10⁻⁶ F
Inductance in series LCR circuit, L = 1 Henry
RMS value of voltage, ε_rms = 50 V
Frequency of source, f = 50/ $π$ Hz
Reactance of the inductor (X_L) is given by,
X_L= $ω L$ = 2 $π$ fL
$\Rightarrow$ X_L = 2 $\times π \times \frac{50}{π} \times 1$ = 100 $Ω$
Reactance of the capacitance $(X_{C})$ is given by,
$X_{C} = \frac{1}{ω C}$ = $\frac{1}{2 π f C}$
$\Rightarrow$ X_C = $\frac{1}{2 π \times \frac{50}{π} \times 20 \times 10^{- 6}}$
$\Rightarrow$ X_C = $500$ $Ω$
(a) Impedance of an LCR circuit $(Z)$ is given by,
$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}} \Rightarrow Z = \sqrt{{(300)}^{2} + {(500 - 100)}^{2}} \Rightarrow Z = \sqrt{{(300)}^{2} + {(400)}^{2}} \Rightarrow Z = 500$
RMS value of current $(I_{r m s})$ is given by,
$I_{r m s} = \frac{ε_{r m s}}{Z}$
$\Rightarrow$ I_rms = $\frac{50}{500}$
$\Rightarrow$ I_rms = $0.1 A$
(b) Potential across the capacitor $(V_{C})$ is given by,
V_C = I_rms × X_C
$\Rightarrow$ V_C = 0.1 × 500 = 50 V
Potential difference across the resistor $(V_{R})$ is given by,
V_{R =}I_rms × R
$\Rightarrow$ V_R = 0.1 × 300 = 30 V
Potential difference across the inductor $(V_{L})$ is given by,
V_L = I_rms × X_L
$\Rightarrow$ V_L= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied

Page No 330:

Answer:

Given:
Resistance in the LCR circuit, R = 300 Ω
Capacitance in the LCR circuit, C = 20 μF = 20 × 10⁻⁶ F
Inductance in the LCR circuit, L = 1 henry
Net impedance of the LCR circuit, Z = 500 ohm
RMS value of voltage, $ε_{rms}$ = 50 V
RMS value of current, I_rms = 0.1 A
Peak current $(I_{0})$ is given by,
$I_{0} = \frac{E_{rms} \sqrt{2}}{Z} = \frac{50 \times 1.414}{500} = 0.1414 A$
Electrical energy stored in capacitor $(U_{C})$ is given by,
$U_{C} = \frac{1}{2} C V^{2}$
$\Rightarrow U_{C} = \frac{1}{2} \times 20 \times 10^{- 6} \times 50 \times 50 \Rightarrow U_{C} = 25 \times 10^{- 3} J = 25 mJ$
Magnetic field energy stored in the coil $(U_{L})$ is given by,
$U_{L} = \frac{1}{2} L I_{0}^{2}$
$\Rightarrow U_{L} = \frac{1}{2} \times 1 \times {(0.1414)}^{2} \Rightarrow U_{L} \approx 5 \times 10^{- 3} J \Rightarrow U_{L} = 5 mJ$

Page No 330:

Answer:

Given:
Inductance of inductor, L = 2.0 H
Capacitance of capacitor, C = 18 μF
Resistance of resistor, R = 10 kΩ
Voltage of AC source, E = 20 V
(a) In an LCR circuit, current is maximum when reactance is minimum, which occurs at resonance, i.e. when capacitive reactance becomes equal to the inductive reactance,i.e.
X_L = X_C
$\Rightarrow ω L = \frac{1}{ω C} \Rightarrow ω^{2} = \frac{1}{L C} = \frac{1}{2 \times 18 \times 10^{- 6}} \Rightarrow ω^{2} = \frac{10^{6}}{36} \Rightarrow ω = \frac{10^{3}}{6} \Rightarrow 2 π f = \frac{10^{3}}{6} \Rightarrow f = \frac{1000}{6 \times 2 π} = 26.539 Hz = 27 Hz$

(b) At resonance, reactance is minimum.
     Minimum Reactance, Z = R
     Maximum current (I) is given by,
   $I = \frac{E}{R} \Rightarrow I = \frac{20}{10 \times 10^{3}}$
   $\Rightarrow I = \frac{2 A}{10^{3}} = 2 mA$

Page No 330:

Answer:

RMS value of voltage, E_rms = 24 V
Internal resistance of battery, r = 4 Ω
RMS value of current, I_rms = 6 A
Reactance $(R)$ is given by,
$R = \frac{E}{I_{rms}} \Rightarrow R = \frac{24}{6} = 4 Ω$
Let R' be the total resistance of the circuit. Then,
   R' = R + r
$\Rightarrow$ R' = 4 Ω + 4 Ω
$\Rightarrow$ R' = 8 Ω

Current, I = $\frac{12}{8} A$
                = 1.5 A

Page No 330:

Answer:

Here,
Input voltage to the filter, Vi = 10 × 10⁻³V
Resistance of the circuit, R = 1 × 10³ Ω
Capacitance of the circuit, C = 10 × 10⁻⁹ F
(a) When frequency, f = 10 kHz
A low pass filter consists of resistance and capacitance. Voltage across the capacitor is taken as the output.
     Capacitive reactance $(X_{C})$ is given by,
$X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}$
   $\Rightarrow X_{C} = \frac{1}{2 π \times 10 \times 10^{3} \times 10 \times 10^{- 9}} \Rightarrow X_{C} = \frac{1}{2 π \times 10^{- 4}} \Rightarrow X_{C} = \frac{10^{4}}{2 π} = \frac{5000}{π} Ω$
Net impedence of the resistance-capacitance circuit (Z) is given by,
   $Z = \sqrt{R^{2} + X_{C}^{2}} \Rightarrow Z = \sqrt{(1 + 10^{3}) + {(5000 / π)}^{2}} \Rightarrow Z = \sqrt{10^{6} + {(5000 / π)}^{2}}$
Current (I₀) is given by,
$I_{0} = \frac{V_{i}}{Z} \Rightarrow I_{0} = \frac{10 \times 10^{- 3}}{\sqrt{10^{6} + {(5000 / π)}^{2}}}$
Output across the capacitor $(V_{0})$ is given by,
$V_{0} = \frac{10^{2}}{\sqrt{10^{5} + {(50 / π)}^{2}}} \times \frac{500}{π} \Rightarrow V_{0} = 1.6124 V = 1.6 mV$
(b)When frequency, f = 1 MHz = $1 \times$ 10⁶ Hz
Capacitive reactance $(X_{C})$ is given by,
$X_{c} = \frac{1}{ω C} \Rightarrow X_{C} = \frac{1}{2 π f C} \Rightarrow X_{C} = \frac{1}{2 π \times 10^{6} \times 10^{- 9} \times 10} \Rightarrow X_{C} = \frac{1}{2 π \times 10^{- 2}} \Rightarrow X_{C} = \frac{100}{2 π} \Rightarrow X_{C} = \frac{500}{π} Ω Total impedence (Z) = \sqrt{R^{2} + {X_{C}}^{2}} \Rightarrow Z = \sqrt{{(10^{3})}^{2} + {(50 / π)}^{2}} Current (I_{0}) = \frac{V_{1}}{Z} \Rightarrow I_{0} = \frac{10 \times 10^{- 3}}{\sqrt{10^{6} + {(50 / π)}^{2}}} Output voltage (V_{0}) = I_{0} X_{C} \Rightarrow V_{0} = \frac{10^{- 2}}{\sqrt{10^{5} + {(50 / π)}^{2}}} \times \frac{50}{π} \Rightarrow V_{0} = 0.16 mV$

(c) When frequency, f = 10 MHz = 10 × 10⁶ Hz = 10⁷ Hz
       Capacitive reactance $(X_{C})$ is given by,
       $X_{c} = \frac{1}{ω C} \Rightarrow X_{C} = \frac{1}{2 π f C} \Rightarrow X_{C} = \frac{1}{2 π \times 10^{7} \times 10 \times 10^{- 9}} \Rightarrow X_{C} = \frac{5}{π} Ω Impedence (Z) = \sqrt{R^{2} + X_{c}^{2}} \Rightarrow Z = \sqrt{{(10^{3})}^{2} + {(5 / π)}^{2}} Current (I_{0}) = \frac{V_{1}}{Z} \Rightarrow I_{0} = \frac{10 \times 10^{- 3}}{\sqrt{10^{6} + {(5 / π)}^{2}}} V_{0} = I_{0} X_{C} \Rightarrow V_{0} = \frac{10^{- 2}}{\sqrt{10^{6} + {(5 / π)}^{2}}} \times \frac{5}{π} \Rightarrow V_{o} = 16 μV$

Page No 331:

Answer:

A transformer works on the principle of electromagnetic induction, which is only possible in case of AC.
Hence, when DC (zero frequency) is supplied to it, the primary coil blocks the current supplied to it. Thus, the induced current in the secondary coil is zero. So, the output voltage will be zero.

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