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(A) In a steady state,the capacitor acts as an open circuit,so no current flows through it.First,we find the equivalent resistance of the circuit to d

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) In a steady state,the capacitor acts as an open circuit,so no current flows through it.First,we find the equivalent resistance of the circuit to determine the current flowing from the battery.The $10 \, \Omega$ resistor is in parallel with the capacitor,but since no current flows through the capacitor,the current flows through the $10 \, \Omega$ resistor.The $2 \, \Omega$ and $10 \, \Omega$ resistors are in series,giving a total resistance of $12 \, \Omega$.This combination is in parallel with the $4 \, \Omega$ resistor: $R_p = \frac{12 	imes 4}{12 + 4} = \frac{48}{16} = 3 \, \Omega$.The total resistance of the circuit is $R_{eq} = 6 \, \Omega + 3 \, \Omega = 9 \, \Omega$.The total current from the battery is $I = \frac{V}{R_{eq}} = \frac{72 \, V}{9 \, \Omega} = 8 \, A$.The voltage across the $4 \, \Omega$ resistor (and the parallel combination of $2 \, \Omega$ and $10 \, \Omega$) is $V_p = I 	imes R_p = 8 \, A 	imes 3 \, \Omega = 24 \, V$.The current flowing through the branch containing the $2 \, \Omega$ and $10 \, \Omega$ resistors is $I' = \frac{24 \, V}{2 \, \Omega + 10 \, \Omega} = 2 \, A$.The voltage across the capacitor is the same as the voltage across the $10 \, \Omega$ resistor: $V_c = I' 	imes 10 \, \Omega = 2 \, A 	imes 10 \, \Omega = 20 \, V$.The charge on the capacitor is $q = C 	imes V_c = 10 \, \mu F 	imes 20 \, V = 200 \, \mu C$.

Determine the charge on the capacitor in the following circuit ................ $\mu C$.

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