## An Inductor 20 mH , a capacitor 100 μF and a resistor 50 ohm are connected in series across a source…

Q: An Inductor 20 mH , a capacitor 100 μF and a resistor 50 ohm are connected in series across a source of emf V = 10 sin314 t . The power loss in circuit is

(a) 2.74 W

(a) 1.13 W

(a) 0.43 W

(a) 0.79 W

Ans: (d)

Sol: R = 50 Ω ;

XL = ω L = 314 x 20 x 10-3 = 6.28 ohm

$\displaystyle X_C = \frac{1}{\omega C}$

$\displaystyle X_C = \frac{1}{314 \times 100 \times 10^{-6}}$

XC = 31.85 ohm

$\displaystyle Z^2 = R^2 + (X_C – X_L)^2$

$\displaystyle Z^2 = (50)^2 + (25.5)^2 = 3150.25$

$\displaystyle P = \frac{1}{2} V_0 I_0 cos\phi$

$\displaystyle P = \frac{1}{2} V_0 (\frac{V_0}{Z}) (\frac{R}{Z})$

$\displaystyle P = \frac{V_0^2 R}{2 Z^2}$

$\displaystyle P = \frac{10^2 \times 50}{2 \times 3150.25}$

P = 0.79 W

## In L.C.R. series resonant circuit, an inductance of 10 mH , a resistance of 30 ohms and a capacitance of 0.1 mF are…

Q: In L.C.R. series resonant circuit, an inductance of 10 mH , a resistance of 30 ohms and a capacitance of 0.1 mF are connected with an A.C. the e.m.f. of the A.C. source is given a E = 20 cos⁡ωt . What will be the amplitude of current at a frequency 20% less than the resonant frequency?

(a) 0.66 A

(b) 6.6 A

(c) 66 A

(d) 6.6 × 10-3 A

Ans: (a)

Sol: Here, L = 10 mH = 10-2 H ,R = 30 Ω,

C = 0.1 mF = 10-4 F

Resonance frequency $\displaystyle \omega = \frac{1}{\sqrt{L C}}$

$\displaystyle \omega = \frac{1}{\sqrt{10^{-2}\times 10^{-4}}}$

Actual frequency, ω’ = 80%×1000 = 800 rad/s

XL = ω’ L = 800×10-2 = 8 ohm

$\displaystyle X_C = \frac{1}{\omega’ } = \frac{1}{800 \times 10^{-4}}$

= 12 .5 ohm

$\displaystyle Z = \sqrt{R^2 + (X_L – X_C)^2}$

$\displaystyle Z = \sqrt{30^2 + (4.5)^2}$

Z = 30.33 ohm

$\displaystyle I_0 = \frac{E_0}{Z} = \frac{20}{30.33}$

= 0.66 A

## A 200 km long telegraph wire has a capacity of 0.014 μF/km. If it carries an alternating current of 50 cycles/s then…

Q: A 200 km long telegraph wire has a capacity of 0.014 μF/km. If it carries an alternating current of 50 cycles/s then what is the value of an inductance required to be connected in series so that impedance is minimum ?

(a) 0.36 mH

(b) 0.56 mH

(c) 0.76 mH

(d) 3.6 H

Ans: (d)

Sol: Total capacitance of telegraph wire,

C = 0.014 × 200 µF = 2.8×10-6 F

ν = 50 c/s, L = ?

Impedance is minimum at resonance, for which

$\displaystyle \nu = \frac{1}{2\pi \sqrt{L C}}$

on squaring ,

$\displaystyle L = \frac{1}{4 \pi^2 \nu^2 C}$

on putting the given values ,

L = 3.6 H

## An LCR circuit contains resistance of 100 ohm and a supply of 200 volt at 300 radian/sec angular frequency…

Q: An LCR circuit contains resistance of 100 ohm and a supply of 200 volt at 300 radian/sec angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by 60°. If on the other hand, the only inductor is taken out, the current leads by 60° with the applied voltage. The current flowing in the circuit is

(a) 1 amp

(b) 1.5 amp

(c) 2 amp

(d) 2.5 amp

Ans: (c)

Sol: Here, R = 100 Ω, Ev = 200 V, ω = 300 radian/s

$\displaystyle tan60 = \frac{X_L}{R}$

$\displaystyle tan60 = \frac{X_C}{R}$

Hence , XL = XC

Resonance occurs .

$\displaystyle I_v = \frac{E_v}{Z} = \frac{E_v}{R}$

$\displaystyle I_v = \frac{200}{100}$

= 2 A

## An electric bulb has a rated power of 50 W at 100 V. If it is used on an AC source of 200 V, 50 Hz…

Q: An electric bulb has a rated power of 50 W at 100 V. If it is used on an AC source of 200 V, 50 Hz, a choke has to be used in series with it. This choke should have an inductance of

(a) 1 mH

(b) 0.1 mH

(c) 0.1 H

(d) 1.1 H

Ans: (d)

Sol: Here, P = 50 W, V = 100 volt

I = P/V=50/100 = 0.5 A

R = V/I= 100/0.5 = 200 Ω

Let L be the inductance of the choke coil

$\displaystyle I_v = \frac{E_v}{Z}$

$\displaystyle Z = \frac{E_v}{I_v}$

$\displaystyle Z = \frac{200}{0.5}$

Z = 400 Ω

$\displaystyle X_L = \sqrt{Z^2 – R^2}$

$\displaystyle X_L = \sqrt{400^2 – 200^2}$

$\displaystyle X_L = 200\sqrt{3}$

$\displaystyle \omega L = 200\sqrt{3}$

$\displaystyle L = \frac{200\sqrt{3}}{\omega}$

$\displaystyle L = \frac{200\sqrt{3}}{2 \pi \nu}$

$\displaystyle L = \frac{200\sqrt{3}}{2 \pi \times 50}$

L = 1.1 H