Electrostatics Ques 186
- 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) $C E$
Show Answer
Answer:
Correct Answer: 186.(b)
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, $Q=C V=\frac{C E r_{2}}{r+r_{2}}$
BETA
