Question

In the figure shown, the current in the $10\text{V}$ battery is close to :

• A. $0.71\text{A}$ from positive to negative terminal
• B. $0.42\text{A}$ from positive to negative terminal
• C. $0.21\text{A}$ from positive to negative terminal
• D. $0.36\text{A}$ from negative to positive terminal

Answer 1

The correct option is C $0.21\text{A}$ from positive to negative terminal

Using Kirchhoff's loop law $\text{ABCD}$,

$-5i-10i_1-2i+20=0$

$\Rightarrow 7i+10i_1=20...\left(i\right)$

Using Kirchhoff's loop law in loop $\text{BEFC}$

$-10-4(i-i_1)+10i_1=0$

$\Rightarrow -4i+14i_1=10...\left(ii\right)$

From eqs $\left(i\right)\&\left(ii\right)$, we get

$i_1=\frac{150}{138}=1.09\text{A};i=1.30\text{A}$

$\therefore i-i_1=0.21\text{A}$

$i-i_1$ is positive it means current flows from positive to negative terminal.

Hence, option $\left(\text{C}\right)$ is correct.