Physics - Capacitor Question with Solution | TestHub

Ans (16.00)

Capacitors in series share the same charge. Equivalent capacitance: $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

Step 1: $\frac{1}{C_{eq}}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\Rightarrow C_{eq}=\frac{8}{3}\mu F$

Step 2: Apply Kirchhoff's law. Let Q be the charge on each capacitor (same in series).

Voltage equation for the first loop (12 V battery and 4 $\mu$F capacitor): $12-\frac{Q}{4}-V_{junction}=0$

Voltage equation for the second loop (6 V battery and 8 $\mu$F capacitor): $V_{junction}-\frac{Q}{8}-6=0$

Step 3: Adding the two equations eliminates $V_{junction}$ : $6=\frac{3}{8}Q\Rightarrow Q=\frac{6\times 8}{3}=16\mu C$

The charge on the 4 $\mu$F capacitor is 16 $\mu$C.