Question 48: Electromagnetic Induction

Question

Two inductors L1 (inductance 1mH, internal resistance 3$$\Omega$$) and L2 (inductance 2 mH, internal resistance 4$$\Omega$$), and a resistor R (resistance 12$$\Omega$$) are all connected in parallel across a 5V battery. The circuit is switched on at time t = 0. The ratio of the maximum to the minimum current (Imax / Imin) drawn from the battery is

Accepted Answer

The circuit is shown in the figure
Initially, right after the circuit is switched on ($ t ightarrow 0^{+} $), the impedance (effective resistance) of inductors $ L_1 $ and $ L_2 $ is extremely high. Therefore, the inductors act as open circuits, and all the current flows through the resistor $ R = 12 \Omega $. According to Kirchhoff's law, the current through the battery at this moment is $ i_{\min} = \frac{V}{R} = \frac{5}{12} \, 	ext{A} $.
As the circuit reaches steady state ($ t ightarrow \infty $), the impedance of the inductors drops to zero, and they function as resistors with their specified internal resistances.
The effective resistance of the circuit is $R_e=(12 \Omega \|$ $4 \Omega)\|3 \Omega=3 \Omega\| 3 \Omega=3 / 2 \Omega$.
In the steady state, the current through the circuit is $ i_{\max} = \frac{V}{R_e} = \frac{5}{\frac{3}{2}} = \frac{10}{3} \, 	ext{A} $.
Therefore, the ratio of the maximum current to the minimum current is:
$ \frac{i_{\max}}{i_{\min}} = \frac{\frac{10}{3}}{\frac{5}{12}} = \frac{10}{3} 	imes \frac{12}{5} = 8 $

Electromagnetic Induction

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