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

Question

The network shown in figure is a part of a complete circuit. If at a certain instant the current 'I' is 5A and is decreasing at the rate of 103As- 1 , then VA-VB is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-fngbrhopzderb0hz.jpg" /> (A) 5V (B) 10V (C) 15V (D) 20V

Tardigrade Admin · Accepted Answer

(d i/d t)=-103A/sec <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-fngbrhopzderjymo.jpg" /> Apply KVL VA-5 × 1-15 V-5 × 10-3 ×((d i/d t))-VB=0 VA-5 × 1-15-5 × 10-3 ×(-103)-VB=0 VA-VB=15 V

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

$5 V$

$10 V$

$15 V$

$20 V$

$\frac{d i}{d t} = - 1 0^{3} A / sec$

Apply KVL
$V_{A} - 5 \times 1 - 15 V - 5 \times 1 0^{- 3} \times (\frac{d i}{d t}) - V_{B} = 0$
$V_{A} - 5 \times 1 - 15 - 5 \times 1 0^{- 3} \times (- 1 0^{3}) - V_{B} = 0$
$V_{A} - V_{B} = 15 V$

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

5V

10V

15V

20V