An ideal choke draws a current of 8 , A when | vedclass

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(A) Resistance of the resistor: $R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.Inductive reactance of the choke at $50 \, Hz$: $X_L = \frac{V}{I} =

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$\frac{15}{\sqrt{2}} \, A$

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(A) Resistance of the resistor: $R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.Inductive reactance of the choke at $50 \, Hz$: $X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.Since $X_L = 2 \pi f L$, we have $12.5 = 2 \pi 	imes 50 	imes L$, which gives $L = \frac{12.5}{100 \pi} = \frac{1}{8 \pi} \, H$.Now, for the new source of $40 \, Hz$, the new inductive reactance is $X_L' = 2 \pi f' L = 2 \pi 	imes 40 	imes \frac{1}{8 \pi} = 10 \, \Omega$.The impedance of the series $RL$ circuit is $Z = \sqrt{R^2 + X_L'^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \Omega$.The current in the circuit is $I = \frac{V'}{Z} = \frac{150}{10\sqrt{2}} = \frac{15}{\sqrt{2}} \, A$.

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