Question:
In the figure shown, the current in the $10 \mathrm{~V}$ battery is close to:
Correct Option: , 3
Solution:
$\mathrm{E}_{\mathrm{eq}}=\frac{20 \times 10}{17}=\frac{200}{17}$
and $\mathrm{R}_{\text {eq }}=\frac{7 \times 10}{17}=\frac{70}{17}$
$\therefore \quad I=\frac{\frac{20}{17}-10}{4+\frac{70}{17}}=0.21 \mathrm{~A}$
from +ve to -ve terminal
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.