## AC Circuits

### Maharashtra Board-Class-12th-Physics-Chapter-13

### Solutions

**Question 1. Choose the correct option**.

**(i) If the rms current in a 50 Hz AC circuit is 5A, the value of the current 1/300 seconds after its value becomes zero is**

**(A) 5\(\sqrt{2}\) A**

**(B) 5\(\sqrt{\frac{3}{2}}\) A**

**(C) \(\frac{5}{6}\) A **

**(D) \(\frac{5}{\sqrt{2}}\)** **A**

(B) 5\(\sqrt{\frac{3}{2}}\) A

**(ii) A resistor of 500 ****Ω ****and an inductance of 0.5 H are in series with an AC source which is given by V = 100****\(\sqrt{2}\)**** sin (1000 t). The power factor of the combination is**

**(A) \(\frac{1}{\sqrt{2}}\)**

**(B) \(\frac{1}{\sqrt{3}}\)**

**(C) 0.5 **

**(D) 0.6**

(A) \(\frac{1}{\sqrt{2}}\)

**(iii) In a circuit L, C & R are connected in series with an alternating voltage of frequency f. the current leads the voltage by 450. The value of C is**

**(A) \(\frac{1}{πf(2πfL-R)}\)**

**(B) ** ** \(\frac{1}{2πf(2πfL-R)}\)**

**(C) \(\frac{1}{πf(2πfL+R)}\)**

**(D) \(\frac{1}{2πf(2πfL+R)}\)**

(B) \(\frac{1}{2πf(2πfL-R)}\)

**(iv) In an AC circuit, e and i are given by e = 150 sin (150t) V and i = 150 sin (150 t + ****π/****3 ) A. the power dissipated in the circuit is**

**(A) 106W **

**(B) 150W**

**(C) 5625W **

**(D) Zero**

(C) 5625W

**(v) In a series LCR circuit the phase difference between the voltage and the current is 450. Then the power factor will be**

**(A) 0.607 **

**(B) 0.707**

**(C) 0.808 **

**(D) 1**

(B) 0.707

**Question 2. Answer in brief.**

**(i) An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance ?**

Capacitive reactance: Xc = \(\frac{1}{2πfC}\)

The capacitive reactance is inversely proportional to the capacitance. The capacitive reactance decreases as the capacitance increases. The current flow increases as capacitive reactance decreases. The brightness is proportional to the magnitude of the current, therefore the brightness of the bulb increases as the current increases. Therefore it can be concluded that the brightness of the bulb increases as the capacitance increases.

**(ii) The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why ? **

For an LR circuit, the impedance, Z_{LR} = \(\sqrt{R^2+X_L^2}\), where X_{L} is the reactance of the inductor.

When a capacitor of capacitance C is added in series with L and R, the impedance,

Z_{LCR} = \(\sqrt{R^2+(X_L-X_C)^2}\) because in the case of an inductor the current lags behind the voltage by a phase angle of π/2 rad while in the case of a capacitor the current leads the voltage by a phase angle of π/2 rad. The decrease in net reactance

decreases the total impedance (Z_{LCR} < Z_{LR}).

**(iii) For very high frequency AC supply, a capacitor behaves like a pure conductor. Why ?**

The reactance of a capacitor is Xc = \(\frac{1}{2πfC}\), where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, Xc is very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.

**(iv) What is wattles current ?**

**Wattless current : **The current that does not lead to energy consumption, hence zero power consumption, is called wattless current.

In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current.

**(v) What is the natural frequency of L C circuit ? What is the reactance of this circuit at this frequency ?**

The natural frequency of LC circuit is \(\frac{1}{2π\sqrt{LC}}\)

where L is the inductance and C is the capacitance.

The reactance of this circuit at this frequency is \(\frac{1}{2πfC-\frac{1}{2πfL}}\) = \(\frac{1}{\frac{2πC}{2π\sqrt{LC}}-\frac{1}{\frac{2πL}{2π\sqrt{LC}}}}\) = \(\frac{1}{\sqrt{\frac{C}{L}}-\sqrt{\frac{C}{L}}}\) = \(\frac{1}{0}\) **= **∞

**Question 3. In a series LR circuit X_{L} = R and power factor of the circuit is P_{1} . When capacitor with capacitance C such that X_{L} = X_{C} is put in series, the power factor becomes P_{2}. Calculate P_{1}/P_{2}.**

For a series LR circuit, power factor,

cos Φ = \(\frac{R}{\sqrt{R^2+X_L^2}}\)

If X_{L} = R, power factor P_{1} = \(\frac{R}{\sqrt{2R^2}}\) = \(\frac{1}{\sqrt{2}}\)

For a series LCR circuit, power factor,

cos Φ = \(\frac{R}{\sqrt{R^2+(X_L-X_C)^2}}\)

If X_{L} = Xc, power factor P_{2} = \(\frac{R}{R}\) = 1

∴ *P _{1}*

**/**\(\frac{1}{\sqrt{2}}\)

*P*_{2}=

**Question 4. When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.**

In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of π/2 rad. Here, for e = e_{0} sin ωt, we have,

i = i_{0} sin(ωt − π/2)

Instantaneous power,

P = ei = (e_{0} sin ωt)[i_{0}(sin ωt cos π/2 − cos ωt sin π/2)]

= − e_{0}i_{0} sin ωt cos ωt (… as cos π/2 =0 and sin π/2 =1)

Average power over one cycle,

P_{av} = \(\frac{\text{work done in one cycle}}{\text{time for one cycle}}\)

= \(\frac{\int_{0}^{T}P\,dt}{T}\) = \(\frac{-\int_{0}^{T}e_0i_0\,sin\,ωt\,cos\,ωt\,dt}{T}\)

= \(-\frac{e_0i_0}{T}\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\)

Now, \(\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\) = \(\frac{1}{2}\int_{0}^{T}sin\,2ωt\,dt\)

= \(\frac{1}{2}[\frac{-cos\,2ωt}{2ω}]_0^T\)

= \(-\frac{1}{4ω}[cos\,2\frac{2π}{T}-cos\,0]\)

= \(-\frac{1}{4ω}[1-1]=0\)

∴ P_{av} = 0

**Question 5. Prove that an ideal capacitor in an AC circuit does not dissipate power.**

In an AC circuit containing only an ideal capacitor, the current i leads the emf e by a phase angle of π/2 rad.

Here, for e = e_{0} sin ωt, we have, i = i_{0} sin(ωt + π/2)

Instantaneous power,

P = ei = (e_{0} sin ωt)[i_{0}(sin ωt cos π/2 + cos ωt sin π/2)]

= e_{0}i_{0} sin ωt cos ωt (… as cos π/2 =0 and sin π/2 =1)

Average power over one cycle,

P_{av} = \(\frac{\text{work done in one cycle}}{\text{time for one cycle}}\)

= \(\frac{\int_{0}^{T}P\,dt}{T}\) = \(\frac{\int_{0}^{T}e_0i_0\,sin\,ωt\,cos\,ωt\,dt}{T}\)

= \(\frac{e_0i_0}{T}\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\)

Now, \(\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\) = \(\frac{1}{2}\int_{0}^{T}sin\,2ωt\,dt\)

= \(\frac{1}{2}[\frac{-cos\,2ωt}{2ω}]_0^T\)

= \(-\frac{1}{4ω}[cos\,2\frac{2π}{T}-cos\,0]\)

= \(\frac{1}{4ω}[1-1]=0\)

∴ P_{av} = 0, i.e., the circuit does not dissipate power.

**Question 6. (a) An emf e = e_{0} sin **

**ω**

*t***applied to a series L - C – R circuit derives a current**

*I*=*I*_{0}sin**ω**

*t***in the circuit. Deduce the expression for the average power dissipated in the circuit. (b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.**

(a) Let *e *= e_{0} sin ωt be the alternating emf applied across the series combination of pure inductor, capacitor and resistor as shown in Fig.

There is a phase difference φ between the applied emf and current given by

*i *= *i*_{0} sin (ω*t *± Φ )

Instantaneous power is given by

*P *= *ei = *(e_{0} sin ωt)[* i*_{0} sin (ω*t *± Φ )]

= (e_{0} sin ωt)[i_{0}(sin ωt cos Φ ± cos ωt sin Φ)]

= e_{0}i_{0} sin ωt (sin ωt cos Φ ± cos ωt sin Φ)

= e_{0}i_{0} cos Φ sin^{2}ωt ± e_{0}i_{0} sin Φ sin ωt cos ωt

Average power over one cycle,

P_{av} = \(\frac{\text{work done in one cycle}}{\text{time for one cycle}}\)

= \(\frac{\int_{0}^{T}P\,dt}{T}\)

** = ** \(\frac{\int_{0}^{T}[e_0i_0cos\,Φ\,sin^2 ωt±e_0i_0sin\,Φ\,sin\,ωt\,cos\,ωt]dt}{T}\)

= \(\frac{e_0i_0}{T}[cos\,Φ\int_{0}^{T}sin^2 ωt±sin\,Φ\int_{0}^{T}sin\,ωt\,cos\,ωt]dt]\)

Now, \(\int_{0}^{T}i^2\,sin^2ωt\,dt\) = \(\int_{0}^{T}\frac{1-cos2ωt}{2}dt\)

= \(\int_{0}^{T}\frac{1}{2}dt-\int_{0}^{T}\frac{cos2ωt}{2}dt\)

= \(\frac{T}{2}-\frac{1}{2}(\frac{sin2ωt}{2ω})_{0}^{T}\)

= \(\frac{T}{2}-\frac{1}{4ω}(sin2ωT-sin0)\)

= \(\frac{T}{2}-\frac{1}{4ω}[sin2(\frac{2π}{T})T-0]\)

= \(\frac{T}{2}-\frac{1}{4ω}[0-0]\) = \(\frac{T}{2}\)

Also

\(\int_{0}^{T}sin\,ωt\,cos\,ωt\,dt\) = \(\frac{1}{2}\int_{0}^{T}sin\,2ωt\,dt\)

= \(\frac{1}{2}[\frac{-cos\,2ωt}{2ω}]_0^T\)

= \(-\frac{1}{4ω}[cos\,2\frac{2π}{T}-cos\,0]\)

= \(\frac{1}{4ω}[1-1]=0\)

Hence ** **P_{av} = \(\frac{e_0i_0}{T}cos\,Φ×\frac{T}{2}\) = \(\frac{e_0i_0}{2}cos\,Φ\) ** **

** ** = \(\frac{e_0}{\sqrt{2}}.\frac{i_0}{\sqrt{2}}cos\,Φ\)

= \(e_{rms}.i_{rms}cos\,Φ\) = \(e_{rms}.i_{rms}\frac{R}{Z}\)

where the impedance Z = \(\sqrt{R^2+(X_L-X_C)^2}\)

(b) P_{av} = \(e_{rms}.i_{rms}cos\,Φ\)

- The factor cos Φ is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than
*e*. It means there is signiﬁcant loss of power during transportation_{rms}.i_{rms}**.**

**Question 7. A device Y is connected across an AC source of emf e = e_{0}sin**

**ω**

*t***. The current through Y is given as**

*i*=*i*_{0}sin(**ω**

*t***+**

**π**

**/2)**

**(a) Identify the device Y and write the expression for its reactance.**

**(b) Draw graphs showing variation of emf and current with time over one cycle of AC for Y.**

**(c) How does the reactance of the device Y vary with the frequency of the AC ? Show graphically**

**(d) Draw the phasor diagram for the device Y.**

(a) The device Y is a capacitor. Its reactance is X_{C }= \(\frac{1}{ωC}\)

where ω is the angular frequency of the applied emf and C is the capacitance of the capacitor.

(b) Graphs showing variation of emf and current with time over one cycle of AC for Y.

(c) X_{C }= \(\frac{1}{ωC}\)= \(\frac{1}{2πfC}\), Thus X_{C} ∝ \(\frac{1}{f}\) where f is the frequency of AC.

Suppose C = \(\frac{1000}{2π}\) pF, ∴ Xc = M Ω

For f = 100 Hz, Xc = 1 x 10^{7} Ω = 10 M Ω;

for f = 200 Hz, Xc = 5 M Ω;

for f = 300 Hz, Xc = 10/3 M Ω;

for f = 400 Hz, Xc = 2.5 M Ω

for f = 500 Hz, Xc = 2 M Ω and so on (1 ΜΩ = 10^{6}Ω).

(d) The phasor representing the peak emf (e_{0}) makes an angle ot in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i_{0}) is turned 90° anticlockwise with respect to the phasor representing emf e_{0}. The projections of these phasors on the vertical axis give instantaneous values of e and i.

**Question 8. Derive an expression for the impedance of an LCR circuit connected to an AC power supply.**

Now let us consider the total opposition offered by a resistor, pure inductor and capacitor connected in series with the alternating source of emf as shown in Fig.

Let a pure resistor *R*, a pure inductance *L *and an ideal capacitor of capacitance *C *be connected in series to a source of alternative emf. As *R*, *L *and *C *are in series, the current at any instant through the three elements has the same amplitude and phase. Let it be represented by *i *= *i*_{0} sin ω*t*.

The voltage across the resistor, e_{R} = Ri, is in phase with the current. The voltage across the inductor, e_{L} = X_{L}i, leads the current by π/2 rad and that across the capacitor, e_{c} = X_{C}i, lags behind the current by π/2 rad. This is shown in the phasor Diagram (Fig.).

From this ﬁgure, e_{0}^{2} = e_{R}^{2} + (e_{L} − e_{C})^{2}

= R^{2}i_{0}^{2} + (X_{L}i_{0} − X_{C}i_{0})^{2} = i_{0}^{2} [R^{2} + (X_{L }− X_{C})^{2}]

∴ e_{0} = \(i_0\sqrt{R^2+(X_L-X_C)^2}\) = i_{0}Z, Where

Z = \(\frac{e_0}{i_0}\) = \(\sqrt{R^2+(X_L-X_C)^2}\) is the effective resistance of the circuit. It is called the impedance.

**Question 9. Compare resistance and reactance.**

**Comparison between resistance and reactance :**

Resistance |
Reactance |

Resistance, which opposes the flow of charges (current), can be found in both DC and AC circuits. | Reactance is a term that only appears in an AC circuit. When an inductor or capacitor is used, it happens. |

In a purely resistive circuit, current and voltage are always in phase. | There is a nonzero phase difference between current and voltage when reactance is not zero. |

Resistance does not depend on the frequency of AC. | Reactance depends on the frequency of AC. In case of an inductor, reactance increases linearly with frequency. In case of a capacitor, reactance decreases as frequency of AC increases; it is inversely proportional to frequency. |

Resistance gives rise to production of Joule heat in a component. | In a circuit with pure reactance, there is no production of heat. |

**Question 10. Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by ****π****/2 in phase.**

Let us connect the source of alternating emf to a circuit containing pure inductor (*L*) only as shown in Fig.

Let us assume that the inductor has negligible resistance. The circuit is therefore a purely inductive circuit.

Suppose the alternating emf supplied is represented by

*e *= e_{0} sin ωt

On closing the key K, an emf is induced in the inductor as the magnetic ﬂux linked with it changes with time. This emf opposes the applied emf and according to the laws of electromagnetic induction by Faraday and Lenz, we have,

e' = \(L\frac{di}{dt}\) .....(1)

where e’ is the induced emf and i is the current through the inductor. To maintain the current, e and e’ must be equal in magnitude and opposite in direction.

According to Kirchh0ff’s voltage law, as the resistance of the inductor is assumed to be zero, we have,

e = −e’ = \(L\frac{di}{dt}\) …….(2)

∴ \(L\frac{di}{dt}\) = e/L = (e_{0} sin ωt)/L

∴ ∫ di = \(\int \frac{e_0\,sin\,ωt}{L}dt\)

∴ i = \(-\frac{e_0}{ωL}cos\,ωt+C\)

where C is the constant of integration. C must be time independent and have the dimension of current. As e oscillates about zero, i also oscillates about zero and hence there cannot be any time independent component of current.

∴ C = 0. ∴ i = \(-\frac{e_0}{ωL}cos\,ωt\) = \(-\frac{e_0}{ωL}sin(\frac{π}{2}-ωt)\) …….(3)

∴ i = \(\frac{e_0}{ωL}sin(ωt-\frac{π}{2})\) … [ as sin(-θ) = − sin θ]

From Eq. (3), *i _{peak} = i_{0}* = \(\frac{e_0}{ωL}\)

∴ i = \(i_0sin(ωt-\frac{π}{2})\)

Comparison of this equation with e = e_{0} sin ωt shows that e leads i by π/2 rad, i.e., the voltage is ahead of current by π/2 rad in phase.

**Question 11. An AC source generating a voltage e = e_{0}sin **

**ω**

*t***is connected to a capacitor of capacitance C. Find the expression for the current**

*i*flowing through it. Plot a graph of*e*and*i*versus wt.Let us consider a capacitor with capacitance *C *connected to an AC source with an emf having instantaneous value *e *= e_{0} sin ωt. This is shown in Fig.

The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V = q/C, ∴ q = CV.

q and V are functions of time, with V = e = e_{0} sin ωt. The instantaneous current in the circuit is

i = dq/dt = \(\frac{d}{dt}(CV)\) = \(C\frac{dV}{dt}\) = \(C\frac{d}{dt}(e_0\,sinωt)\) = ωCe_{0 }cos ωt

∴ i = \(\frac{e_0}{(1/ωC)}sin(ωt+\frac{π}{2})\) = \(i_0\,sin(ωt+\frac{π}{2})\)

Where i_{0} = \(\frac{e_0}{(1/ωC)}\) is the peak value of the current,

From Eq. *e *= e_{0} sin ωt and Eq. \(i=i_0\,sin(ωt+\frac{π}{2})\) we find that in an AC circuit containing a capacitor only, the alternating current *i *leads the alternating emf *e *by phase angle of π/2 radian as shown in above graph.

**Question 12. If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current 1/600 sec. after if was zero ?**

**Given** : f = 50 Hz, i_{rms} = 5A, t = \(\frac{1}{600}\) s

The peak value of the current, i_{0} = i_{rms}\(\sqrt{2}\) = (5)(1.414) = **7.07 A**

i = i_{0 }sin(2πft)

= 7.07 sin [2π(5)\((\frac{1}{600})\)]

= 7.07 sin \(\frac{π}{6}\)

= (7.07)(0.5) = 3.535 A

**3.535 A**** is the required current.**

**Question 13. A light bulb is rated 100W for 220 V AC supply of 50 Hz. Calculate (a) resistance of the bulb. (b) the rms current through the bulb.**

Given : Power (V_{rms }i_{rms}) = 100W, V_{rms} = 220 V, f = 50 Hz

The rms current through the bulb,

i_{rms }= \(\frac{power}{V_{rms}}\) = \(\frac{100}{220}\) = 0.4545 A

The resistance of the bulb, R = \(\frac{V_{rms}}{i_{rms}}\) = \(\frac{220}{100/220}\) = 484 Ω

**Question 14. A 15.0 ****μ****F capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current.**

**Given **: C = 15 μF = 15 x 10^{−}^{6}F, V_{rms} = 220V, f = 50 Hz,

The capacitive reactance = \(\frac{1}{2πfC}\) = \(\frac{1}{2×3.142×50×(15×10^{-6}}\) = 212.2 Ω

i_{rms }= \(\frac{V_{rms}}{capacitive\,reactance}\) = \(\frac{220}{212.2}\) = 1.037 A

i_{peak} = i_{rms}\(\sqrt{2}\) = (1.037)(1.414) = 1.466 A

If the frequency is doubled, the capacitive reactance will be halved and the current will be doubled.

**Question 15. An AC circuit consists of only an inductor of inductance 2H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor (****π ****= 3.142),**

**Given** : L = 2H, i_{0} = 0.25 A, f = 60 Hz, π = 3.142,

ωL = 2πfL = 2(3.142)(60)(2) = 754.192

The effective potential difference across the inductor = ωLi_{rms} = ωL\(\frac{i_0}{\sqrt{2}}\) = \(\frac{(754.1)(0.25)}{1.414}\) = **133.3 V**

**Question 16. Alternating emf of e = 220 sin 100 ****π****t is applied to a circuit containing an inductance of (1/****π****) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?**

**Given** : e = 220sin 100 πt, L = (1/π)H

Comparing e = 220sin 100 πt with e = e_{0} sin ωt, we get

ω = 100π, ∴ ωL = 100π × \(\frac{1}{π}\) = 100 Ω

∴ The instantaneous current through the circuit = i = \(\frac{e_0}{ωL}\) sin(100 πt – π/2)

= sin(100 πt – π/2) = 2.2sin(100 πt – π/2) A

i_{rms} = \(\frac{i_0}{\sqrt{2}}\) = \(\frac{2.2}{1.414}\) = **1.556 A** is the reading of the AC galvanometer connected in the circuit.

**Question 17. A 25 ****μ****F capacitor, a 0.10 H inductor and a 25 ****Ω ****resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?**

Data: C = 25 μF = 25 x 10^{−}^{6} F, L = 0.10 H, R = 25 Ω, e = 310 sin (314t) [volt]

Comparing e = 310 sin (314 t) with e = e_{0} sin (2πft), we get, the frequency of the alternating emf as f = \(\frac{314}{2π}\) = \(\frac{314}{2×3.14}\) = 50 Hz

Reactance = \(|ωL-\frac{1}{ωC}|\) = \(|2πfL-\frac{1}{2πfC}|\) = \(|(2)(3.142)(50)(0.10)-\frac{1}{(2)(3.142)(50)(25×10^{-6}}|\) = |31.4 − 127.4| = 96 Ω

Z^{2} = R^{2} + \((ωL-\frac{1}{ωC})^2\) = 25^{2} + 96^{2 }= 9841 Ω^{2}

∴ Impedance Z = \(\sqrt{9841}\) = 99.2 Ω

Peak current i_{0} = e_{0}/Z = 310/99.2 A

∴ i_{rms} = \(\frac{i_0}{\sqrt{2}}\) = (\frac{310}{(1.414)(99.2)}\) = 2.21 A

cos Φ = R/Z = 25/99.2 = 0.2520

∴ Phase angle Φ = cos^{−1 }**(**0.2520) = 75.40^{0 }= 1.316 rad

**Question 18. A capacitor of 100 ****μ****F, a coil of resistance 50****Ω ****and an inductance 0.5 H are connected in series with a 110 V - 50 Hz source. Calculate the rms value of current in the circuit.**

**Given** : C = 100 µF = 100 x 10^{−}^{6} F = 10^{−}^{4} F, R = 50 Ω, L = 0.5H, f = 50 Hz,

V_{rms} = 110 V

∴ ωL = 2πfL = 2(3.142)(50)(0.5) = 157.152

And \(\frac{1}{ωC}\) = \(\frac{1}{2πfC}\) = \(\frac{1}{(2)(3.142)(50)(10^{-4}}\) = 31.83 Ω

Z^{2} = R^{2} + \((ωL-\frac{1}{ωC})^2\) = (50)^{2} + (157.1 - 31.83)^{2 }= 2500 + 15700 = 18200 Ω^{2}

∴ Impedance, Z = \(\sqrt{18200}\) Ω = 134.9 Ω

The rms value of the current in the circuit,

i_{rms }= \(\frac{V_{rms}}{Z}\) = \(\frac{110}{134.9}\) = **0.8154 A**

**Question 19. Find the capacity of a capacitor which when put in series with a 10 ****Ω ****resistor makes the power factor equal to 0.5. Assume an 80 V – 100 Hz AC supply.**

**Given** : R = 10 Ω, power factor = 0.5, f = 100 Hz,

Power factor = \(\frac{1}{2πfCR}\)

0.5 = \(\frac{1}{(2)(3.142)(100)C(10)}\)

∴ C = \(\frac{1}{3.142×10^3}\) = 3.182 × 10^{−}^{4 }F

This is the capacity of the capacitor.

**Question 20. Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.**

**Given** : f = 50 Hz, i = \(\frac{i_0}{\sqrt{2}}\) ∴ \(\frac{i}{i_0}\) = \(\frac{1}{\sqrt{2}}\)

i = i_{0} sin ωt, ∴ sin ωt = \(\frac{i}{i_0}\) = \(\frac{1}{\sqrt{2}}\)

∴ ωt = π/4 rad

∴ 2πft = π/4

∴ t = \(\frac{1}{8f}\) = \(\frac{1}{8×50}\) = 2.5 × 10^{−}^{3 }s

**2.5 ****× 10**^{−}^{3 }**s**** is the required time**

**Question 21. Calculate the value of capacity in picofarad, which will make 101.4 micro henry inductance to oscillate with frequency of one megahertz.**

**Given**: f = 10^{6} Hz, L = 101.4 x 10^{−}^{6} H

f_{r} = \(\frac{1}{2π\sqrt{LC}}\),

∴ f_{r}^{2} = \(\frac{1}{4π^2LC}\) ,

∴ C = \(\frac{1}{4π^2f_r^2L}\) = \(\frac{1}{4(3.142)^2(10^6)^2(101.4×10^{-6}}\) = 2.497 × 10^{−}^{10} F

= 249.7 × 10^{−}^{12} F = 249.7 picofarad

This is the value of the capacity.

**Question 22. A 10 ****μ****F capacitor is charged to a 25 volt of potential. The battery is disconnected and a pure 100 m H coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil.**

**Given** : C = 10 µF = 10 x 10^{−}^{6} F = 10** ^{−5}** F, L = 100 mH = 100 × 10

^{−}^{3}H = 10

^{−}^{1}H,

V = 25 V

The energy stored in the electric field in the capacitor = \(\frac{1}{2}\)Cv^{2}

The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li^{2 }

Here, \(\frac{1}{2}\)Cv^{2 }= \(\frac{1}{2}\)Li^{2}

∴ i^{2} = \(\frac{C}{L}V^2\) = \(\frac{10^{-5}}{10^{-1}}(25)^2\) = 25^{2} x 10^{−4}

∴ i = 25 x 10^{−2 }**A = ****0.25 A**

This is the maximum current in the coil.

**Question 23. A 100 ****μ****F capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.**

**Given** : C = 100 µF = 100 x 10^{−}^{6} F = 10^{−}^{4} F, V = 50 V, i = 5A

The energy stored in the electric field in the capacitor = \(\frac{1}{2}\)CV^{2}

The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li^{2 }

Here, \(\frac{1}{2}\)CV^{2 }= \(\frac{1}{2}\)Li^{2}

∴ L = \(C(\frac{V}{i})^2\) = \(10^{-4}(\frac{50}{5})^2\) = 10** ^{−2 }**H

This is the value of the inductance.

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