Discussion Forum

Two inductors  L₁ (inductance lmH, internal resistance 3Ω) and k (inductance 2 mH, internal resistance 4 Ω), and a resistor R (resistance 12 Ω) 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 (I_max , I_min ) drawn from the battery is

Answer:

Explanation:

$I_{max}=\frac{\epsilon}{R}=\frac{5}{12}A$

(Initially t=0)

$I_{min}=\frac{\epsilon}{R_{eq}}=\epsilon \left(\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{R}\right)$

(finally in steady state)

$=5\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{12}\right)=\frac{10}{3}A$

$\frac{I_{max}}{I_{min}}=8$