JEE Main 2017 · 2 Apr · Shift 1 · Physics Q16

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PhysicsSingle correct+4 / -1Question 16 of 90

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :

Three parallel branches: the top has the cell E with internal resistance r; the middle has the capacitor C in series with r₁; the bottom has r₂.

CE\dfrac{r_1}{(r_2+r)}

CE\dfrac{r_2}{(r+r_2)}

CE\dfrac{r_1}{(r_1+r)}

Answer and worked solution

Official answer: C

In steady state no current flows through the capacitor branch. The current $\dfrac E{r+r_2}$ flows through $r_2$ , so the capacitor (in parallel with $r_2$ , with $r_1$ carrying no current) has voltage $\dfrac{Er_2}{r+r_2}$ :
$Q=CE\frac{r_2}{r+r_2}.$

Topics:RC circuits Kirchhoff's laws

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JEE Main 2017 · 2 Apr · Shift 1, Physics question 16