Question

Determine the potential difference between the points $C$ and $D$ in the given figure.

• A. 3.6V
• B. 7.2V
• C. 12V
• D. 10.8V

Answer 1

Answer: (A)

Step 1 : Main current I = $\frac{12}{(1+6+3)}$ = 1.2 A
Potential difference across AB= 12 - 1.2 $\times$ l = 10.8 V $(terminalvoltage{V}_{tmnl})$
Potential difference across AD= 6 $\times$ 1.2 = 7.2 V
Step 2 : . Potential difference across AC = $\frac{q}{C}$ ...$(i)$
Capacitors are in series $\therefore$ $q$ is same or $q=V_{tmnl}\times C_{eq}$
Here, $\frac{1}{C_{eq}}=\frac{1}{1}+\frac{1}{2}=\frac{2+1}{2}$ or $C_{eq}=\frac{2}{3}\mu$F
$\therefore$ $q=10.8\times \frac{2}{3}\times 10^{-6}=7.2\times 10^{-6}$ C
Putting in equation $(i)$, Potential difference across AC = $\frac{7.2\times 10^{-6}}{2\times 10^{-6}}$ = 3.6 V
Step 3 : Potential difference between C and D is 7.2 - 3.6 = 3.6 V
SHort CuT Terminal voltage of 10.8V is available across series combination of capacitors of l$\mu$F and 2$\mu$F in the ratio of 2 : 1 giving 3.6V across 2$\mu$F capacitor. Similarly terminal voltage of 10.8 V is available to series combination of 3$\Omega$ and 6$\Omega$ in the ratio of 1 : 2 giving 7.2 V across 6$\Omega$ resistor