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 circuit is in steady state with switch S closed. Now the switch S is opened. Just after the switch S is opened. the current through resistance `R_(1)` is

Options

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

Answer

A

Explanation

At the given instant p.d across capacitor is 20 volt. Hence the current through `R_(1)` at the required instant of time is `I = (40 - 20)/(R_(1)) =1 amp`.