Consider the circuit shown in the figure, where C_1 = 4.00 mu F, C_2 = 1.00 mu F

Question

Consider the circuit shown in the figure, where C_1 = 4.00 mu F, C_2 = 1.00 mu F, and delta V = 18.0 V. Capacitor C_1 is first charged by closing switch S_1. Switch S_1 is then opened, and the charged capacitor is connected to the uncharged capacitor by closing S_2. q_1 and q_2 charges for capacitor with C_1 and C_2, respectively. Calculate the final charge on each capacitor. q_1 = 7.6 mu C and q_2 = 4.4 mu C q_1 = 4.4 mu C and q_2 = 7.6 mu C q_1 = 57.6 mu C and q_2 = 14.4 mu C q_1 = 14.4muC and q_2 =57.6 mu C q_1 = 10.0 mu C and q_2 = 10.0 mu C

Explanation / Answer

Apply Q=CV

When S1 is closed ; Qo=4*18 =72 uC

When S1 is open and S2 is closed voltage gets equalled at last. Take it as V across each capacitor;

Apply Q=CV for C1; Q1=4*V ; for Q2; Q2=1*V

Since the total chrge is conserved, Q1+Q2= Qo

5V=72 ; V=14.4

So, Q1= 57.6 uC ; Q2=14.4 uC

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