Question

Two capacitors of $3\mu F$ and $6\mu F$ are connected in series and a potential difference of $900V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the combination is

Answer 1

Let $C1=3\mu F,C2=6\mu F$

$C1andC2$ are connected in series with potential difference $V=900V$ acros the combination

Thus for the series combination Charge on both capacitor is same

$\therefore C_1{V}_1=C_2{V}_2=Q...1$

$\therefore \frac{V_1}{V_2}=\frac{C_2}{C_1}=2\mu C...2$

$also{V}_1+V_2=900V...3$

From $2and3$

$V_2=300Vand{V}_1=600V$

Therefore charge on each capacitor $Q=300\times 6=600\times 3=1800\mu C$

When these capacitors are disconnected and reconnected in parallel the charge is distributed and the potential difference is same

$\therefore \frac{Q_1}{C_1}=\frac{Q_2}{C_2}=V...4$

$\therefore \frac{Q_1}{Q_2}=\frac{C_1}{C_2}=\frac{1}{2}...5$

$and{Q}_1+Q_2=Q=1800\mu C...6$

From $5and6$

$Q_1=600\mu Cand{Q}_2=1200\mu C$

Therefore the potential difference across the combination is $V=\frac{600}{3}=\frac{1200}{6}=200V$

Ans is $200V$