Question

I₄(0) ... current through R₄ just after switch is closed ... = 0.04196 A

A circuit is constructed with four resistors, one inductor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 48 Ω, R3 = 102 Ω and R4 = 136 Ω. The inductance is L = 342 mH and the battery voltage is V = 12 The positive terminal of the battery is indicated with a +sign. R3

I₄(∞) ... current through R₄ after switch has been closed for a very long time ... = 0.05539 A

I_L(∞) ... current through inductor after switch has been closed for a very long time ... = 0.03767 A

V_L,max(closed) ... maximum voltage across inductor during the time when switch is closed ... = 4.28 V

V_L,max(open) ... maximum voltage across inductor during the time when switch is open ... = 5.65 V

PLEASE HELP SOLVE:

Answer 1

The inductor works as a voltage source when the switch is opened as magnetic energy gets stored inside it.
Current across inductor when the switch is opened :

, R is the resistance across which the current flows from the inductor through R₂ and R₃.

So, R = R2+R3 = 48+102ohm = 150ohm, putting in values for t = $5.7*10^{-3}s$ , we get,

$I_t = \frac{5.6}{150}*e^{-\frac{5.7*10^{-3}*150}{342*10^{-3}}} = 3.064*10^{-3}A$

As I_t is in the opposite direction of I_L , current flows anticlockwise, so I₃ is at the same direction(Positive).