In the circuit diagram shown,E = 5 , V, r = 1 | vedclass

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(A) In a steady state,capacitors act as open circuits. Therefore,no current flows through the branches containing capacitors.The circuit simplifies to

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$6$

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(A) In a steady state,capacitors act as open circuits. Therefore,no current flows through the branches containing capacitors.The circuit simplifies to a single loop containing the battery $E$,internal resistance $r$,and resistor $R_2$ in series.The current in the circuit is $I = \frac{E}{R_2 + r} = \frac{5 \, V}{4 \, \Omega + 1 \, \Omega} = 1 \, A$.The potential difference across $R_2$ is $V_{R2} = I 	imes R_2 = 1 \, A 	imes 4 \, \Omega = 4 \, V$.Since the capacitor branches are in parallel with $R_2$,the potential difference across each branch containing capacitors is also $4 \, V$.For the top branch,the two capacitors $C$ are in series,so their equivalent capacitance is $C_{eq1} = \frac{C 	imes C}{C + C} = \frac{C}{2} = \frac{3 \, \mu F}{2} = 1.5 \, \mu F$.The charge on the capacitors in the top branch is $q_1 = C_{eq1} 	imes V = 1.5 \, \mu F 	imes 4 \, V = 6 \, \mu C$.Similarly,for the bottom branch,the equivalent capacitance is $C_{eq2} = 1.5 \, \mu F$,and the charge is $q_2 = 1.5 \, \mu F 	imes 4 \, V = 6 \, \mu C$.Thus,the magnitude of the charge on each capacitor plate is $6 \, \mu C$.

In the circuit diagram shown,$E = 5 \, V, r = 1 \, \Omega, R_2 = 4 \, \Omega, R_1 = R_3 = 1 \, \Omega$ and $C = 3 \, \mu F$. The magnitude of the charge on each capacitor plate is......$\mu C$.

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