When a coil is connected to an AC supply of f | vedclass

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(C) Given: Frequency $f = 50 \, Hz$, Current $I = 4 \, A$, Power $P = 240 \, W$, Voltage $V = 100 \, V$.The power consumed by the coil is given by $P

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$L=\frac{1}{5 \pi} \, H$

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(C) Given: Frequency $f = 50 \, Hz$, Current $I = 4 \, A$, Power $P = 240 \, W$, Voltage $V = 100 \, V$.The power consumed by the coil is given by $P = I^2 R$.Substituting the values: $240 = (4)^2 	imes R \Rightarrow 240 = 16R \Rightarrow R = 15 \, \Omega$.The impedance $Z$ of the coil is given by $Z = \frac{V}{I} = \frac{100}{4} = 25 \, \Omega$.We know that $Z^2 = R^2 + X_L^2$, where $X_L$ is the inductive reactance.$(25)^2 = (15)^2 + X_L^2 \Rightarrow 625 = 225 + X_L^2 \Rightarrow X_L^2 = 400 \Rightarrow X_L = 20 \, \Omega$.Since $X_L = 2 \pi f L$, we have $20 = 2 \pi (50) L$.$20 = 100 \pi L \Rightarrow L = \frac{20}{100 \pi} = \frac{1}{5 \pi} \, H$.

When a coil is connected to an $AC$ supply of frequency $50 \, Hz$, a current of $4 \, A$ flows in it and it consumes $240 \, W$ power. If the potential difference across the coil is $100 \, V$, then the inductance value of the coil is

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