Question

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

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

Answer 1

Answer: (B)

0.9 A

In steady state, the capacitor draws no current from cell. So, no current flow in the capacitor branch and it is removed from the circuit and corresponding circuit is shown in figure.
Total resistance of the circuit is $R_t=2.8+2||3=2.8+\frac{2\times 3}{2+3}=2.8+1.2=4\Omega$
Current in the circuit is $I=\frac{6}{4}=1.5A$
If $I_1$ and $I_2$ are the current through $2\Omega$ and $3\Omega$ respectively, then $I=I_1+I_2...\left(1\right)$
As $2\Omega$ and $3\Omega$ are in parallel so $V_{2\Omega }=V_{3\Omega }$ or $2I_1=3I_2\Rightarrow I_2=\left(2/3\right)I_1$
now, (1) becomes, $1.5=I_1+\left(2/3\right)I_1=\frac{5I_1}{3}$
or $I_1=\frac{1.5\times 3}{5}=0.9A$