Resistances of 6 Ω each are connected in the manner shown in figure with the current 0.5 A as shown in figure, the potential difference, VP-VQ is

Question

Resistances of 6 Ω each are connected in the manner shown in figure with the current 0.5 A as shown in figure, the potential difference, VP-VQ is (A) 3 V (B) 5 V (C) 4 V (D) 3.9 V

Tardigrade Admin · Accepted Answer

As Vp-Vq =((6/3)+(12 × 6/12+6))(0.5) =(2+4) × 0.5 V =6 × 0.5 V =3 V

Q. Resistances of $6 Ω$ 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

3 V

5 V

4 V

3.9 V

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

Q. Resistances of 6Ω each are connected in the manner shown in figure with the current 0.5A as shown in figure, the potential difference, VP​−VQ​ is

3 V

5 V

4 V

3.9 V

As Vp​−Vq​=(36​+12+612×6​)(0.5) =(2+4)×0.5V=6×0.5V=3V

Q. Resistances of $6 Ω$ 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

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