91Ó°ÊÓ

When the switch is just closed, the inductor opposes the sudden change in the current. Initially current in the circuit is zero and it will remain zero just after the switch is closed. Hence, inductorLact as an open circuit. Now,R1andR3becomes in series connection. So, the current in the circuit is given by E=13A

The voltage across R1 is given by${\mathrm{V}}_{\mathrm{R}1}=33.33\mathrm{V}$

The voltage across R3 is given by ${\mathrm{V}}_{\mathrm{R}3}=16.66\mathrm{V}$

The voltage across R2 is ${\mathrm{V}}_{\mathrm{R}2}=0$

When the switch is closed for a long time, the current in the circuit reached the equilibrium state and the inductor acts as a short-circuited. The equivalent resistance of the circuit is given by Req=R1+R2×R3R2+R3=100+75×5075+50=130 Ω

The total current in the circuit is given by$\mathrm{l}=513\mathrm{A}$

The voltage V1 is ${\mathrm{V}}_1=38.40\mathrm{V}$,

The current through the resistor R₁ is${\mathrm{i}}_1=0.385\mathrm{A}$

The inductor acts as a short circuit. So, the voltage across it is zero.${\mathrm{V}}_2=0$

The current through R₂ is, ${\mathrm{i}}_2=0.154\mathrm{A}$

The voltage ${\mathrm{V}}_3$is,${\mathrm{V}}_3=11.54\mathrm{V}$

The current through ammeter A₃ is${\mathrm{i}}_3=0.231\mathrm{A}$

The voltage across R3 is${\mathrm{V}}_4=11.54\mathrm{V}$