JEE Main Solved Paper 2017 Question 28
Question: In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be: [JEE Main Solved Paper-2017]
Options:
A) $ CE\frac{r _2}{(r+r _2)} $
B) $ CE\frac{r _1}{(r _1+r)} $
C) CE
D) $ CE\frac{r _1}{(r _2+r)} $
Show Answer
Answer:
Correct Answer: A
Solution:
- It steady state, current through AB = 0
$ \Rightarrow $ $ V _{AB}=V _{CD} $
$ \Rightarrow $ $ V _{AB}=( \frac{\varepsilon }{r+r _2}\times r _2 )-V _{CD} $
$ \Rightarrow $ $ Q _{c}=CV _{AB} $ $ =CE( \frac{r _2}{r+r _2} ) $
