Question

In the circuit shown, the current is $2\Omega$ resistance will be

• A. 1.25A
• B. 1.5A
• C. 1.8A
• D. 0.9A

Answer 1

Answer: (D)

0.9A

In steady state, no current flow in the capacitor branch. Thus the current will flow through resistors 2.8, 3 and 2 .

So , total resistance of the circuit is $R_{eq}=2.8+2||3=2.8+\frac{2\times 3}{2+3}=2.8+1.2=4\Omega$

Thus, current , $I=\frac{V}{R_{eq}}=6/4=1.5A$

As resistor 2 and 3 are in parallel so current I will distribute in their indirect ratio.

Thus, current through resistor $2\Omega$ is $i_2=\frac{3}{2+3}I=\left(3/5\right)\left(1.5\right)=0.9A$