Question

A circuit contains a battery, a capacitor, and a resistor as shown. The capacitor is initially uncharged, and the switch is closed at time $t=0.$ When the charge on the capacitor is at $75\%$ of its maximum possible value, what is the voltage drop across the resistor (in $\text{V}$) (integer only)

Answer 1

Given, $\text{At some instant charge on capacitor Q}=75\%\text{of}{Q}_{max}$

$\because Q_{max}=C{V}_{max}[\because \Delta V_{max}=12\text{V}]$

Potential across the capacitor at the same instant when $Q=75\%\text{of}{Q}_{max}$ is,

$\Delta V_C=\frac{Q}{C}=\frac{0.75Q_{max}}{C}=0.75\Delta V_{max}$

$=0.75\times 12=9\text{V}$

Using KVL we get,

$\Delta V_C+\Delta V_R–E=0$

$\Rightarrow \Delta V_R=12–9=3\text{V}$