The network shown in figure is part of a complete circuit. If at a certain instant, the current I is 5A and decreasing at a rate 103As- 1 then VB-VA is:

Question

The network shown in figure is part of a complete circuit. If at a certain instant, the current I is 5A and decreasing at a rate 103As- 1 then VB-VA is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-ck6v6v32wtixz6mq.jpg" /> (A) 20V (B) 15V (C) 10V (D) 5V

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ck6v6v32wtixo66t.jpg" /> From KVL A to B VA-(.1× 5.)+15+(5 × (10)- 3)(10)3=VB VB-VA=+15volt

Q. The network shown in figure is part of a complete circuit. If at a certain instant, the current $I$ is $5 A$ and decreasing at a rate $1 0^{3} A s^{- 1}$ then $V_{B} - V_{A}$ is:

$20 V$

$15 V$

$10 V$

$5 V$

From $K V L$ $A$ to $B$
$V_{A} - (1 \times 5) + 15 + (5 \times (10)^{- 3}) (10)^{3} = V_{B}$
$V_{B} - V_{A} = + 15 v o lt$

Q. The network shown in figure is part of a complete circuit. If at a certain instant, the current I is 5A and decreasing at a rate 103As−1 then VB​−VA​ is:

20V

15V

10V

5V