Question

Find the potential difference $V_a-V_b$ between the points a and b shown in each part of the figure.

Answer 1

(a)

Using Kirchhoff voltage loop,

$-12+\frac{2Q}{2\mu F}+\frac{Q_1}{2\mu F}+\frac{Q_1}{4\mu F}=0$------- (1)

In close circuit PQRSP,

$-12+\frac{Q}{2\mu F}+\frac{Q+Q_1}{4\mu F}=0$ --------- (2)

From equn (1) and (2),

=>$2Q+3Q_1=48$--------- (3)

& $3Q-Q_1=48$.

Substituting $Q=4Q_1$

(3)=> $2Q+3Q_1=48$

$8Q_1+3Q_{1}48$

$11Q_1=48Q_1=\frac{48}{11}$

$V_{ab}=\frac{Q_1}{4\mu F}=\frac{48}{11\times 4}=\frac{12}{11}V$

(b)

Potential across a and b is (12-12)v=0

(c)

In this figure, the left and the right branch is symmetry. So the current go to the branch 'ab' fro both sides are opposite and equal. Hence it cancels out.

Thus net charge, Q= 0.$\therefore V=\frac{Q}{C}=\frac{0}{C}=0$

$\therefore V_{ab}=0$

(d)

The net potential, V=$\frac{Netcharge}{Netcapacitance}=\frac{C_1{V}_1+C_2{V}_2+C_3{V}_3}{7}$

V=$\frac{24+24+24}{7}=\frac{72}{7}=10.28V$