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 2017] |
A) \[CE\frac{{{r}_{2}}}{(r+{{r}_{2}})}\]
B) \[CE\frac{{{r}_{1}}}{({{r}_{1}}+r)}\]
C) CE
D) \[CE\frac{{{r}_{1}}}{({{r}_{2}}+r)}\]
Correct Answer: A
Solution :
[a] It steady state, current through AB = 0 |
\[\Rightarrow \]\[{{V}_{AB}}={{V}_{CD}}\] |
\[\Rightarrow \]\[{{V}_{AB}}=\left( \frac{\varepsilon }{r+{{r}_{2}}}\times {{r}_{2}} \right)-{{V}_{CD}}\] |
\[\Rightarrow \]\[{{Q}_{c}}=C{{V}_{AB}}\] |
\[=CE\left( \frac{{{r}_{2}}}{r+{{r}_{2}}} \right)\] |
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