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Question

(A) The rated current of the bulb is $i = \frac{P}{V} = \frac{60}{10} = 6\, A$.In an $RL$ series circuit,the supply voltage $V$ is related to the volt

Vedclass · Accepted Answer

$0.052\, H$

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(A) The rated current of the bulb is $i = \frac{P}{V} = \frac{60}{10} = 6\, A$.In an $RL$ series circuit,the supply voltage $V$ is related to the voltage across the bulb $V_R$ and the voltage across the inductor $V_L$ by $V = \sqrt{V_R^2 + V_L^2}$.Given $V = 100\, V$ and $V_R = 10\, V$,we have $100^2 = 10^2 + V_L^2$.$V_L^2 = 10000 - 100 = 9900$.$V_L = \sqrt{9900} \approx 99.5\, V$.The inductive reactance is $X_L = \frac{V_L}{i} = \frac{99.5}{6} \approx 16.58\, \Omega$.Since $X_L = 2\pi f L$,we have $L = \frac{X_L}{2\pi f} = \frac{16.58}{2 	imes 3.14 	imes 50} = \frac{16.58}{314} \approx 0.0528\, H$.Thus,the required self-inductance is approximately $0.052\, H$.

$A$ $10\, V, 60\, W$ bulb is to be connected to a $100\, V, 50\, Hz$ $AC$ source. To operate the bulb at its rated power,an induction coil is connected in series with it. Calculate the required self-inductance of the coil.

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