Question

The current $I_1$ (in A) flowing through $1\Omega$ resistor in the above circuit is :

• A. $0.25$
• B. $0.2$
• C. $0.5$
• D. $0.4$

Answer 1

The correct option is B $0.2$

Refer image .1

Potential difference across the top most branch is $1V$

Refer image.2

Equivalent resistance of the top most branch is given by the combinations of the $1\Omega$ and the $2\Omega$ resistor. The parallel combination of $1\Omega$ resistor give $0.5\Omega$

Refer image.3

$\therefore$ The equivalent circuit would be

Refer image.4

$R_{total}=2+0.5$

$=2.5\Omega$

$I_{total}=\frac{V}{R_{total}}=\frac{1}{2.5}=\frac{2}{5}=0.4A$

Now, using current division,

Refer image.5

$V=I_1{R}_1=I_2{R}_2$

$T=I_1+I_2$

$\therefore I_1{R}_1=(I-I_1)R_2$

$I_1(R_1+R_2)=I{R}_2$

$I_1=\frac{I{R}_2}{R_1+R_2}=\frac{0.4\times 1}{1+1}=0.2A$

Option (B) is correct.