A current of 4 , A flows in a coil when conne | vedclass

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(B) $1$. When connected to a $12 \, V$ d.c. source, the coil acts as a pure resistor $R$. Using Ohm's law, $R = V/I = 12 \, V / 4 \, A = 3 \, \Omega$.

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$80$

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(B) $1$. When connected to a $12 \, V$ d.c. source, the coil acts as a pure resistor $R$. Using Ohm's law, $R = V/I = 12 \, V / 4 \, A = 3 \, \Omega$.$2$. When connected to an a.c. source, the impedance $Z$ of the $LR$ circuit is $Z = V/I_{ac} = 12 \, V / 2.4 \, A = 5 \, \Omega$.$3$. The impedance of an $LR$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$, where $X_L = 2\pi fL$ is the inductive reactance.$4$. Substituting the values: $5 = \sqrt{3^2 + X_L^2} \implies 25 = 9 + X_L^2 \implies X_L^2 = 16 \implies X_L = 4 \, \Omega$.$5$. Since $X_L = 2\pi fL$, we have $4 = 2\pi 	imes (25/\pi) 	imes L$.$6$. Simplifying: $4 = 50L \implies L = 4/50 \, H = 0.08 \, H = 80 \, mH$.

$A$ current of $4 \, A$ flows in a coil when connected to a $12 \, V$ d.c. source. If the same coil is connected to a $12 \, V, (25/\pi) \, Hz$ a.c. source, a current of $2.4 \, A$ flows in the circuit. The inductance of the coil is: (in $ \, mH$)

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