Two inductor coils of self-inductance 3,H and | vedclass

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Question

(B) In the steady state,the inductors act as ideal wires (zero resistance). The total current $I$ flowing through the circuit is $I = \frac{V}{R} = \f

Vedclass · Accepted Answer

$\frac{1}{100}$

Vedclass · Answer

(B) In the steady state,the inductors act as ideal wires (zero resistance). The total current $I$ flowing through the circuit is $I = \frac{V}{R} = \frac{10\,V}{10\,\Omega} = 1\,A$.Since the inductors are in parallel,the voltage across each is $0\,V$ in steady state,but they carry currents determined by their parallel configuration. However,since they are connected directly across the battery in this specific circuit diagram,the steady state current through each inductor is determined by the resistance in their respective branches. Since the inductors have zero resistance,the current through them would theoretically be infinite if there were no other resistance. Looking at the circuit,the battery and resistor are in series with the parallel combination of $L_1$ and $L_2$. Thus,the total current $I = 1\,A$ flows through the battery. Since inductors have zero resistance,in steady state,they act as short circuits. The entire current $I = 1\,A$ will flow through the inductors. The current divides inversely proportional to inductance: $I_1 = I 	imes \frac{L_2}{L_1+L_2} = 1 	imes \frac{6}{3+6} = \frac{6}{9} = \frac{2}{3}\,A$ and $I_2 = I 	imes \frac{L_1}{L_1+L_2} = 1 	imes \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3}\,A$.Total energy stored $U = \frac{1}{2} L_1 I_1^2 + \frac{1}{2} L_2 I_2^2 = \frac{1}{2} 	imes 3 	imes (\frac{2}{3})^2 + \frac{1}{2} 	imes 6 	imes (\frac{1}{3})^2 = \frac{1}{2} 	imes 3 	imes \frac{4}{9} + \frac{1}{2} 	imes 6 	imes \frac{1}{9} = \frac{2}{3} + \frac{1}{3} = 1\,J$.Heat developed in resistance $H = I^2 R t = (1)^2 	imes 10 	imes 10 = 100\,J$.Ratio = $\frac{U}{H} = \frac{1}{100}$.

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