Question

Resistances of $6\, \Omega$ each are connected in the manner shown in figure with the current $0.5\, A$ as shown in figure, the potential difference, $V_{P}-V_{Q}$ is

• A. 3 V
• B. 5 V
• C. 4 V
• D. 3.9 V

Answer 1

Answer: (A)

As $V_{p}-V_{q} =\left(\frac{6}{3}+\frac{12 \times 6}{12+6}\right)(0.5)$
$=(2+4) \times 0.5\, V =6 \times 0.5\, V =3\, V $