Physics - Capacitor Question with Solution | TestHub

The given circuit consists of a single loop with two batteries and two capacitors connected in series.

From the circuit diagram, the batteries are connected with opposing polarities. The net voltage in the circuit is:

$V_{\text{net}}=20-4=16\text{ V}$

For capacitors in series, the potential difference across a capacitor is inversely proportional to its capacitance. The potential difference $V_1$ across the $C_1=3\mu \text{F}$ capacitor is:

$V_1=V_{\text{net}}\times \frac{C_2}{C_1+C_2}$

Substituting the given values, $C_1=3\mu \text{F}$ and $C_2=5\mu \text{F}$:

$V_1=16\times \frac{5}{3+5}$

$V_1=16\times \frac{5}{8}=10\text{ V}$

Answer: 10 V