Question

In the circuit shown, the current through the 5 O resistor is

• A. $\frac{8}{3}$ A
• B. $\frac{9}{13}$ A
• C. $\frac{4}{13}$ A
• D. $\frac{1}{3}$ A

Answer 1

Answer: (D)

$\frac{1}{3}$ A

Answer 2

Applying Kirchhoff?s second law for closed loop AEFBA, we get $\quad$ $\quad$ -(I + I) ? 5 - I ? 2 + 2 = 0 or $\quad$ $\quad$ 7I + 5I = 2 $\quad\quad\quad\quad$ ... (i) Again, applying Kirchhoff?s second law for a closed loop DEFCD, we get $\quad$ $\quad$ -(I + I) ? 5 - I ? 2 + 2 = 0 or $\quad$ $\quad$ 5I + 7I = 2 $\quad\quad\quad\quad$ ... (ii) On solving, we get $I_{1}=\frac{1}{6} A$ and $I_{2}=\frac{1}{6} A$ $I=I_{1}+I_{2}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3} A\quad\quad\quad$