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

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 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

An AC voltage source described by V = 10 cos (π/2)t is connected to a 1μF capacitor as shown ….

Q: An AC voltage source described by V = 10 cos (π/2)t is connected to a 1μF capacitor as shown in figure. The key K is closed at t = 0 s. The time (t > 0) after which the magnitude of current I reaches its maximum value for the first time is

Numerical

(a) 1s

(b) 2s

(c) 3s

(d) 4s

Ans: (a)

Sol: As circuit containing capacitance only hence

$ \displaystyle I = I_0 cos(\frac{\pi}{2}t + \frac{\pi}{2})$

$\displaystyle I = -I_0 sin(\frac{\pi}{2}t ) $

It will be maximum after time

$\displaystyle t = \frac{T}{4} = \frac{2\pi}{\omega \times 4} $

$ \displaystyle t = \frac{2\pi}{(\pi/2) \times 4} $

t = 1 sec

In the circuit shown in figure, the power consumed is ….

Q: In the circuit shown in figure, the power consumed is

Numerical

(a) zero

(b) $\displaystyle \frac{V_0^2}{2 R} $

(c) $\displaystyle \frac{V_0^2 R}{2(R^2 + \omega^2 L^2) } $

(d) None of these

Ans: (c)

Sol: $ \displaystyle P = V_{rms} I_{rms} cos\phi $

$ \displaystyle P = V_{rms} \frac{V_{rms}}{Z} \frac{R}{Z} $

$ \displaystyle = \frac{V_{rms}^2 R}{Z^2} $

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

$ \displaystyle = \frac{V_0^2 R}{2 (R^2 + \omega^2 L^2)} $