Question

Three identical capacitors are connected in three different configurations as shown in column II. Match the two columns.

Answer 1

As three capacitors are identical so $C_1=C_2=C_3=C$.

1) $C_{eq}=\frac{C}{3},q_{eq}=C_{eq}{V}_{ab}=\frac{C{V}_{ab}}{3}$

here, $q_1=q_2=q_{eq}=\frac{C{V}_{ab}}{3}$ (as they are in series)

and $V_1=V_2=\frac{q_{eq}}{C}=\frac{V_{ab}}{3}$

2) $C_{eq}=3C$

here, $q_1=C_{eq}{V}_{ab}=3C{V}_{ab}$ similarly $q_2=3C{V}_{ab}$

and $V_1=V_2=V_{ab}$ (as they are in parallel)

3) $C_{eq}=\frac{C(C+C)}{C+(C+C)}=\frac{2}{3}C,q_{eq}=C_{eq}{V}_{ab}=\frac{2}{3}C{V}_{ab}$

here $q_1=q_{eq}=\frac{2}{3}C{V}_{ab}$ and $q_2=\frac{C}{C+C}{q}_{eq}=\frac{1}{3}C{V}_{ab}$

$V_1=\frac{q_1}{C}=\frac{2}{3}{V}_{ab}$ and $V_2=\frac{q_2}{C}=\frac{1}{3}{V}_{ab}$

4) $C_{eq}=\frac{C(C+C)}{C+(C+C)}=\frac{2}{3}C,q_{eq}=C_{eq}{V}_{ab}=\frac{2}{3}C{V}_{ab}$

here $q_2=q_{eq}=\frac{2}{3}C{V}_{ab}$ and $q_1=\frac{C}{C+C}{q}_{eq}=\frac{1}{3}C{V}_{ab}$

$V_2=\frac{q_1}{C}=\frac{2}{3}{V}_{ab}$ and $V_1=\frac{q_2}{C}=\frac{1}{3}{V}_{ab}$

Ans: $A\to 2;B\to 1,3;C\to 2;D\to 1,4.$