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

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(B) The total potential difference in the circuit is $V_{net} = 24 \, V - 12 \, V = 12 \, V$.Since the capacitors are connected in series,this potenti

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$-8$

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(B) The total potential difference in the circuit is $V_{net} = 24 \, V - 12 \, V = 12 \, V$.Since the capacitors are connected in series,this potential difference is divided between the two capacitors $C_1$ and $C_2$ in the inverse ratio of their capacitances.The potential difference across capacitor $C_1$ is given by:$V_1 = \frac{C_2}{C_1 + C_2} 	imes V_{net} = \frac{4}{2 + 4} 	imes 12 = \frac{4}{6} 	imes 12 = 8 \, V$.Looking at the circuit,the plate of capacitor $C_1$ connected towards point $B$ is at a higher potential due to the $24 \, V$ battery. Therefore,the potential at $B$ is higher than at $A$ by $8 \, V$.Thus,$V_B - V_A = 8 \, V$,which implies $V_A - V_B = -8 \, V$.

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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