question_answer

In the given circuit shown, below 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, non 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 circuit is shown in the figure. \[i=\frac{E}{R}=\frac{6}{4}\,=1.5A\] Let current \[{{i}_{1}}\] flows through \[2\Omega \] resistance \[2\times {{i}_{1}}=(i-{{i}_{1}})\times 3\] \[2{{i}_{1}}=(1.5-{{i}_{1}})\times 3\] \[5{{i}_{1}}=4.5\] \[{{i}_{1}}=0.9A\]