In the given circuit,the potential difference | vedclass

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(A) The total voltage $V$ is given by $V^2 = V_R^2 + V_L^2$,where $V_R = 54 \, V$ and $V = 120 \, V$.$V_L = \sqrt{V^2 - V_R^2} = \sqrt{120^2 - 54^2} =

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(A) The total voltage $V$ is given by $V^2 = V_R^2 + V_L^2$,where $V_R = 54 \, V$ and $V = 120 \, V$.$V_L = \sqrt{V^2 - V_R^2} = \sqrt{120^2 - 54^2} = \sqrt{14400 - 2916} = \sqrt{11484} \approx 107.16 \, V$.The power consumed by the resistor is $P = V_R \cdot I$,so $I = P / V_R = 16 / 54 \, A$.The voltage across the inductor is $V_L = I \cdot X_L = I \cdot (2 \pi f L)$.Substituting the values: $107.16 = (16 / 54) \cdot 2 \cdot \pi \cdot 60 \cdot L$.$107.16 = (16 / 54) \cdot 377 \cdot L$.$L = (107.16 \cdot 54) / (16 \cdot 377) \approx 5786.64 / 6032 \approx 0.96 \, H$.Rounding to the nearest integer,$L \approx 1 \, H$.

In the given circuit,the potential difference across the resistance is $54 \, V$ and the power consumed by it is $16 \, W$. If the $AC$ frequency is $60 \, Hz$,find the value of $L$ in $H$.

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