Q.A capacitor of capacitance C1 is charged up to potential V and then connected in parallel to an uncharged capacitor of capacitance C2. The final potential difference across each capacitor will be

In series, all capacitors carry the same charge. C_eq = 10/17 μF, Q = 100/17 μC, and V₂ = Q/(2 μF) = 50/17 V. Option (c).

Step 1 — equivalent series capacitance:

1/C_eq = 1/1 + 1/2 + 1/5 = (10 + 5 + 2)/10 = 17/10, so C_eq = 10/17 μF.

Step 2 — common charge (series capacitors share the same charge):

Q = C_eq × V = (10/17 μF)(10 V) = 100/17 μC.

Step 3 — potential difference across the 2·0 μF capacitor: …

Each capacitor is 6 uF.

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For capacitors in parallel, the total capacitance is the sum of individual capacitances. Here, $C_{\text{eq}} = 10\ \mu\text{F}$. Energy stored is $\frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} \times 10 \times 10^{-6} \times (100)^2 = 0.05\ \text{J} = 5.0 \times 10^{-2}\ \text{J}$. The correct option is (D).

The direct A-C capacitor is in parallel with the series A-B-C path (which reduces to $2\,\mu F$), giving a total equivalent capacitance of $6\,\mu F$ between A and C.

Working

Three $4\,\mu F$ capacitors form a triangle with vertices A, B, C: one directly between A and C, one between A and B, and one between B and C.

Between the terminals A and C, there are two parallel paths:

1. The capacitor directly connecting A to C: $C_1 = 4\,\mu F$.
2. The path A $\to$ B $\to$ C, formed by the A-B and B-C capacitors in series: …

Two 4 μF capacitors in series (2 μF) placed in parallel with the third 4 μF capacitor gives 2 + 4 = 6 μF.

Each capacitor has $C = 4\ \mu F$. Check the given option: two in series, then that combination in parallel with the third.

Series combination of two $4\ \mu F$ capacitors:

$\dfrac{1}{C_s} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2} \implies C_s = 2\ \mu F$

This series pair is now connected in parallel with the third $4\ \mu F$ capacitor:

$C_{eff} = C_s + 4 = 2 + 4 = 6\ \mu F$

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$C_{parallel} = 3C$ and $C_{series} = C/3$, so the ratio is $9 : 1$: option (A).

For three equal capacitors each of capacitance $C$:

In parallel, capacitances add:

$C_p = C + C + C = 3C.$

In series, reciprocals add: …

In a parallel combination, both capacitors share the same voltage. Since $Q = CV$, the charge on each is directly proportional to its capacitance, so the ratio $Q_1 : Q_2 = C_1 : C_2$. The answer is $\boxed{\dfrac{C_1}{C_2}}$, option (A).

Series capacitors share equal charge, so V is inversely proportional to C; the 2μF takes 800 V.

Series combination: same charge Q on both plates.

$Q = C_1 V_1 = C_2 V_2$, so $\dfrac{V_1}{V_2} = \dfrac{C_2}{C_1} = \dfrac{4}{2} = 2.$

Thus $V_1 = 2V_2$ (V₁ across the 2μF). With $V_1 + V_2 = 1200$ V:

$2V_2 + V_2 = 1200 \Rightarrow 3V_2 = 1200 \Rightarrow V_2 = 400$ V (across 4μF), and $V_1 = 800$ V (across 2μF).

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5μF = (two 2μF in parallel = 4μF) in parallel with (two 2μF in series = 1μF) → 4 capacitors.

We want 5μF from 2μF units.

• Two 2μF in series: $\dfrac{2\times2}{2+2}=1\,\mu F$.
• Two 2μF in parallel: $2+2=4\,\mu F$.
• Put the 4μF branch in parallel with the 1μF branch: $4+1 = 5\,\mu F$. …

Charge is conserved when the charged capacitor is connected to the uncharged one; sharing it between the two parallel capacitances gives the common final potential.

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