The network shown in the 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 103 A / s, then VB-VA is

Question

The network shown in the 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 103 A / s, then VB-VA is (A) 5 V (B) 10 V (C) 15 V (D) 20 V

Tardigrade Admin · Accepted Answer

By using Kirchhoff's voltage law, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5c0e0f51b194e802141e07eb7dfb7724-.png" /> VA-i R+E-L (d i/d t)=VB ⇒ VB-VA=-5 × 1+15 -5 × 10-3 ×(-103)=15 V

Q. The network shown in the 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 $1 0^{3} A / s$ , then $V_{B} - V_{A}$ is

5 V

10 V

15 V

20 V

By using Kirchhoff's voltage law,

$V_{A} - i R + E - L \frac{d i}{d t} = V_{B}$
$\Rightarrow V_{B} - V_{A} = - 5 \times 1 + 15$
$- 5 \times 1 0^{- 3} \times (- 1 0^{3}) = 15 V$

Q. The network shown in the 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 103A/s, then VB​−VA​ is

5 V

10 V

15 V

20 V

By using Kirchhoff's voltage law, VA​−iR+E−Ldtdi​=VB​ ⇒VB​−VA​=−5×1+15 −5×10−3×(−103)=15V

Q. The network shown in the 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 $1 0^{3} A / s$ , then $V_{B} - V_{A}$ is

By using Kirchhoff's voltage law,

$V_{A} - i R + E - L \frac{d i}{d t} = V_{B}$
$\Rightarrow V_{B} - V_{A} = - 5 \times 1 + 15$
$- 5 \times 1 0^{- 3} \times (- 1 0^{3}) = 15 V$