Two inductors L i (inductance 1   mH , internal resistance 3   Ω ) and L 2 (inductance 2   mH , internal resistance 4   Ω ), and a resistor R (resistance 12   Ω ) are all connected in parallel across a 5   V battery. The circuit is switched on at time t = 0 . The ratio of the maximum current I m a x I m i n drawn from the battery is:

Step-by-step solution

Initally, the inductances will offer infinite resistance and the current will be maximum. I m a x = ε R = 5 12 A (Initially at t = 0) Upon reaching steady state, the inductances will act as pure resistors and hence the current will be minimum. I m i n = ε R e q = ε 1 r 1 + 1 r 2 + 1 R (Finally in steady state) = 5 1 3 + 1 4 + 1 12 = 10 3 A I m a x I m i n = 8