Question

In the circuit shown, switch S is kept closed and the circuit is in steady state.
(a) Find reading of the ideal voltmeter
(b) Now the switch is opened. Find the reading of the voltmeter immedi- ately after the switch is opened.
(c) Fid the heat dissipated in resistance R after the switch is opened.

Answer 1

Correct Answer - (a) E
(b) 2E with polarity reversed
(c) `(1)/(2)E^(2)((L)/(R^(2))+C)`

Comments

Respected sir please give the full solution

Answer 2

(a) There is no current through capacitors, voltmeters and inductors are zero resistance, when the circuit is in steady state. The effective circuit is as shown..

V_E = V_G and V_G - V_D = E

\(\therefore\) Reading of voltmeter = E

(b) The current through inductors and the voltages across the capacitors cannot change immediately. Current, before the switch is opened, is I = \(\frac ER\) through both inductors.

After opening the switch, the current in R and the inductor between B and D must be same (since voltmeter does not conduct). It implies that current through R is still I and V_G - V_D = RI = E

The current I in GD will loop through GDBG. The current I in the inductor between A and G must loop through AGEA.

\(\therefore\) V_GE = 3R.I = 3E

\(\therefore\) V_D - V_E = 2E

(c) When switch was opened the capacitor between A and E was uncharged. The circuit is effectively two disjoint loops- BGD and AGE

Energy stored in L and C gets dissipated in R.

\(\therefore U_R = \frac 12 LI^2 + \frac 12 CE^2\) 

\(= \frac 12 L\left(\frac ER\right)^2 + \frac 12 CE^2 \)

\(=\frac 12 E^2\left(\frac L{R^2} + C\right)\)