Question

In the figure shown, the current in the 10 V battery is close to

• A. $0.21A$ from positive to the negative terminal
• B. $0.36A$ from negative to the positive terminal
• C. $0.42A$ from positive to the negative terminal
• D. $0.71A$ from positive to the negative terminal

Answer 1

The correct option is A

$0.21A$ from positive to the negative terminal

Step1: Assume the current in branches as shown in figure:

Apply KVL in loop 1,

$\begin{array}{l}+20-5i_1-10(i_1+i_2)-2i_1=0\\ 20-5i_1-10i_1-10i_2-2i_1=0\\ -17i_1-10i_2=-20\\ 17i_1+10i_2=20....\left(1\right)\end{array}$

Step2: Apply KVL in loop 2,

$\begin{array}{l}+10-10(i_1+i_2)-4i_2=0\\ 10-10i_1-10i_2-4i_2=0\\ -10i_1-14i_2=-10\\ 5i_1+7i_2=5...\left(2\right)\end{array}$

Multiply equation (1) by 5 and equation (2) by 17 then subtract the equation (1) from equation (2), we get

$\begin{array}{l}85i_1+50i_2=100\\ 85i_1+119i_2=85\\ (-)(-)(-)\\ \overline{0-69i_2=15}\end{array}$

$\begin{array}{c}{i}_2=-\frac{15}{69}\\ =-.2173\text{A}\end{array}$

It means the direction of the current is from the positive to the negative terminal of the battery inside it.

$\therefore i_2=0.21\text{\hspace{0.17em}A}$

Hence, the correct option is (A).