First slide

Power in a.c. circuit

Question

The self-inductance of a choke coil is 10 mH. When it is connected with a 10 V DC source, then the loss of power is 20 watt. When it is connected with 10 volt AC source loss of power is 10 watt. The frequency of AC source will be

Moderate

A

50 Hz
B

60 Hz
C

80 Hz
D

100 Hz

Solution

With DC : $P = \frac{V^{2}}{R} \Rightarrow R = \frac{(10)^{2}}{20} = 5 Ω$
With AC : $P = \frac{V_{rms}^{2} R}{Z^{2}} \Rightarrow Z^{2} = \frac{(10)^{2} \times 5}{10} = 50 Ω^{2}$
Also $Z^{2} = R^{2} + 4 π^{2} v^{2} L^{2}$
$\Rightarrow 50 = (5)^{2} + 4 (3 .14)^{2} v^{2} {(10 \times 10^{- 3})}^{2} \Rightarrow v = 80 Hz$

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