Electrostatics 7 Question 4
4. In the given circuit diagram, when the current reaches steady state in the circuit, the charge on the capacitor of capacitance $C$ will be
(2017 Main)
(a) $C E \frac{r _1}{\left(r _2+r\right)}$
(b) $C E \frac{r _2}{\left(r+r _2\right)}$
(c) $C E \frac{r _1}{\left(r _1+r\right)}$
(d) $C E$
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Solution:
- In steady state no current flows through the capacitor
So, the current in circuit $I=\frac{E}{r+r _2}$
$\because$ Potential drop across capacitor $=$ Potential drop across $r _2$
$$ =I r _2=\frac{E r _2}{r+r _2} $$
$\therefore$ Stored charge of capacitor, $Q=C V=\frac{C E r _2}{r+r _2}$
