Question

In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the $2\Omega$ resistor is :

• A. $0.6A$
• B. $1.2A$
• C. $0.9A$
• D. $1.5A$

Answer 1

The correct option is C $0.9A$

In the steady state, the capacitor acts as an open to the circuit.

Thus equivalent resistance of the circuit becomes,

$2.8\Omega +\frac{2\times 3}{2+3}\Omega =4\Omega$

Thus current extracted from the battery=$\frac{6V}{4\Omega }=1.5A$

Thus current that goes into the $2\Omega$ resistance=$\frac{3}{2+3}\times 1.5A=0.9A$