The circuit consists of two resistors (of resistance `R_(1) = 20 Omega` and `R_(2) = 10 Omega`), a capacitor (of capacitance `C =10 muF`) and two ideal cells. In the circuit shown the capacitor is in steady state and the switch S is open

The current through the resistor `R_(2)` just after the switch S is closed is :

Options

A. 1 ampere
B. 2 ampere
C. 3 ampere
D. 4 ampere

Answer

C

Explanation

In steady statem before the switch S is closed, potential difference across capacitor is 40 volts. Just after switch S is closed, charge and hence potential difference across the capacitor does not change apreciably. So, the potential difference across `R_(2) ` is `40 - 10 = 30` volt. The current through `R_(2)` is 3 ampere.