Two capacitors C_1 = 2 mu F and C_2 = 4 mu F | vedclass

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(B) The net electromotive force $(EMF)$ in the circuit is $E_{net} = 24 \, V - 12 \, V = 12 \, V$.Since the capacitors are connected in series,the pot

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(B) The net electromotive force $(EMF)$ in the circuit is $E_{net} = 24 \, V - 12 \, V = 12 \, V$.Since the capacitors are connected in series,the potential difference $V = 12 \, V$ is divided between them in the inverse ratio of their capacitances.The potential difference across capacitor $C_1$ is given by $V_1 = \frac{C_2}{C_1 + C_2} 	imes V$.Substituting the values,$V_1 = \frac{4 \, \mu F}{2 \, \mu F + 4 \, \mu F} 	imes 12 \, V = \frac{4}{6} 	imes 12 \, V = 8 \, V$.Looking at the circuit,the positive terminal of the $24 \, V$ battery is connected to the right plate of $C_1$,making the potential at $B$ higher than at $A$ across the capacitor $C_1$. Thus,$V_B - V_A = 8 \, V$.Therefore,the potential difference $(V_A - V_B) = -8 \, V$.

Two capacitors $C_1 = 2 \mu F$ and $C_2 = 4 \mu F$ are connected in a circuit as shown in the figure. The potential difference $(V_A - V_B)$ is....$V$

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