question_answer

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

A)  0.6 A

B)  1.2 A

C)  0.9 A

D)  1.5 A

Correct Answer: C

Solution :

In the steady state, no current flows through the branch containing the capacitor. So, the equivalent circuit will be of the form as shown below : The effective resistance of the circuit is \[R=\frac{2\times 3}{2+3}+2.8\] \[=1.2+2.8=4\Omega \] The current through the ciruit is \[i=\frac{E}{R}=\frac{6}{4}=1.5A\] Let current \[{{i}_{1}}\]flows through \[2\Omega \]resistance. \[\therefore \] \[2\times {{i}_{1}}=(i-{{i}_{1}})\times 3\] \[\Rightarrow \] \[2{{i}_{1}}=(1.5-{{i}_{1}})\times 3\] \[\Rightarrow \] \[2{{i}_{1}}=4.5-3{{i}_{1}}\] \[\Rightarrow \] \[5{{i}_{1}}=4.5\] \[\therefore \] \[{{i}_{1}}=0.9A\]