In the given circuit, if point C is connected | vedclass

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(C) The circuit consists of a $5 \, \mu F$ capacitor in series with a parallel combination of capacitors. First, calculate the equivalent capacitance

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$500$

Vedclass · Answer

(C) The circuit consists of a $5 \, \mu F$ capacitor in series with a parallel combination of capacitors. First, calculate the equivalent capacitance of the parallel part: $C_p = 10 \, \mu F + 10 \, \mu F = 20 \, \mu F$ (Wait, looking at the diagram, the top branch has two $10 \, \mu F$ capacitors in series, which is $5 \, \mu F$. Then this is in parallel with another $10 \, \mu F$ capacitor. So $C_{eq} = 5 \, \mu F + 10 \, \mu F = 15 \, \mu F$). The circuit simplifies to a $5 \, \mu F$ capacitor in series with a $15 \, \mu F$ capacitor. Let $V_A = 2000 \, V$ and $V_C = 0 \, V$ (since it is earthed). The potential at $B$ is given by the voltage divider rule for capacitors: $V_B = V_A 	imes \frac{C_1}{C_1 + C_2}$ is incorrect; for capacitors in series, $V_B = V_A 	imes \frac{C_{eq2}}{C_{eq1} + C_{eq2}}$, where $C_{eq1} = 5 \, \mu F$ and $C_{eq2} = 15 \, \mu F$. $V_B = 2000 	imes \frac{5}{5 + 15} = 2000 	imes \frac{5}{20} = 2000 	imes \frac{1}{4} = 500 \, V$.

In the given circuit, if point $C$ is connected to the earth and a potential of $+2000 \, V$ is applied to point $A$, the potential at point $B$ is.....$V$.

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