Question

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

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

Answer 1

Answer: (C)

$E _{ eq }=\frac{20 \times 10}{17}=\frac{200}{17}$
and $R _{ eq }=\frac{7 \times 10}{17}=\frac{70}{17}$


$\therefore \quad I=\frac{\frac{20}{17}-10}{4+\frac{70}{17}}=0.21 \,A$
from +ve to -ve terminal