Find Vx-Vy just after switch is closed?

Question

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/16b6ed820b81b9e0c24f7c79d46d51ed-.png" /> Find Vx-Vy just after switch is closed? (A) 10 V (B) -10 V (C) 5 V (D) 0 V

Tardigrade Admin · Accepted Answer

Pd. across 2 μ F (1) V A - V x =(30/6) × 4 Pd. across 4 μ F (2) V A - V y =(30/6) × 2 by (2)-(1) ( V A - V y )-( V A - V x ) =(30/6) × 2-(30/6) × 4 V x - V y =-10 volt

Q.
Find $V_{x} - V_{y}$ just after switch is closed?

10 V

-10 V

5 V

0 V

$P d$ . across $2 μ F$
(1) $V_{A} - V_{x} = \frac{30}{6} \times 4$
Pd. across $4 μ F$
(2) $V_{A} - V_{y} = \frac{30}{6} \times 2$
by $(2) - (1)$
$(V_{A} - V_{y}) - (V_{A} - V_{x})$
$= \frac{30}{6} \times 2 - \frac{30}{6} \times 4$
$V_{x} - V_{y} = - 10$ volt

Q. Find Vx​−Vy​ just after switch is closed?

10 V

-10 V

5 V

0 V

Pd. across 2μF (1) VA​−Vx​=630​×4 Pd. across 4μF (2) VA​−Vy​=630​×2 by (2)−(1) (VA​−Vy​)−(VA​−Vx​) =630​×2−630​×4 Vx​−Vy​=−10 volt

Q.
Find $V_{x} - V_{y}$ just after switch is closed?

$P d$ . across $2 μ F$
(1) $V_{A} - V_{x} = \frac{30}{6} \times 4$
Pd. across $4 μ F$
(2) $V_{A} - V_{y} = \frac{30}{6} \times 2$
by $(2) - (1)$
$(V_{A} - V_{y}) - (V_{A} - V_{x})$
$= \frac{30}{6} \times 2 - \frac{30}{6} \times 4$
$V_{x} - V_{y} = - 10$ volt