In the given circuit,find the charge on the c | vedclass

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(B) The circuit is in a steady state,so the capacitors act as open circuits. The current flows through the resistors.Let $V_A = 120 \, V$ and $V_B = 0

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Charge on $C_1$ is $80 \, \mu C$

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(B) The circuit is in a steady state,so the capacitors act as open circuits. The current flows through the resistors.Let $V_A = 120 \, V$ and $V_B = 0 \, V$.The potential at point $a$ is $V_a = 120 - I_1 	imes 1 = 120 - (120 / (1+2)) 	imes 1 = 120 - 40 = 80 \, V$.The potential at point $b$ is $V_b = 120 - I_2 	imes 2 = 120 - (120 / (2+1)) 	imes 2 = 120 - 80 = 40 \, V$.The potential difference across $C_1$ is $V_a - V_b = 80 - 40 = 40 \, V$.The charge on $C_1$ is $q_1 = C_1(V_a - V_b) = 4 \, \mu F 	imes 40 \, V = 160 \, \mu C$.Wait,re-evaluating the potential divider: The total resistance of the upper branch is $1+2+3 = 6 \, \Omega$ and the lower branch is $2+1+3 = 6 \, \Omega$.$V_a = 120 - (120/6) 	imes 1 = 100 \, V$.$V_b = 120 - (120/6) 	imes 2 = 80 \, V$.$V_a - V_b = 100 - 80 = 20 \, V$.$q_1 = 4 \, \mu F 	imes 20 \, V = 80 \, \mu C$.For $C_2$: $V_c = 120 - (120/6) 	imes (1+2) = 120 - 60 = 60 \, V$.$V_d = 120 - (120/6) 	imes (2+1) = 120 - 60 = 60 \, V$.Since $V_c = V_d$,the charge on $C_2$ is $0$.Similarly,$V_e = V_f = 0 \, V$,so the charge on $C_3$ is $0$.

In the given circuit,find the charge on the capacitors.

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