Question

Take the potential of the point B in the figure, to be zero. (a) Find the potentials at the points C and D. (b) If a capacitor is connected between C and D, What charge will appear on the capacitor?

Answer 1

(a) Capacitance on upper branch = $\frac{(4\times 8)}{(4+8)}=\frac{8}{3}\mu F$

and capacitance on the lower branch= $\frac{6\times 3}{6+3}=2\mu F$

(I) The charge on the capacitor $\frac{8}{3}\mu F$ is Q=CV= $\frac{8}{3}\times 50=\frac{400}{3}C$

Therefore the potential difference at $4\mu F$=$\frac{400}{3\times 4}=\frac{100}{3}$

And at

$8\mu F$ = $\frac{400}{3\times 8}=\frac{100}{6}$

The potential difference= $\frac{100}{3}-\frac{100}{6}=50\mu V$

(ii) Effective change at 2

$\mu F=50\times 2=100\mu F$

Therefore potential difference at $3\mu F=\frac{100}{3},$

Potential at 6$\mu F=\frac{100}{6}$

Therefore difference=$\frac{100}{3}-\frac{100}{6}=\frac{50}{3}\mu V$

Therefore pot.diff at C and D=$\frac{50}{3}\mu V$