Electromagnetic Induction and Alternating Current 3 Question 9
10. The network shown in figure is part of a complete circuit. If at a certain instant the current $(I)$ is $5 A$ and is decreasing at a rate of $10^{3} A / s$, then $V _B-V _A=\ldots V$
(1997, 1M)
Show Answer
Solution:
- $\frac{d i}{d t}=10^{3} A / s$
$\therefore$ Induced emf across inductance, $|e|=L \frac{d i}{d t}$
$$ |e|=\left(5 \times 10^{-3}\right)\left(10^{3}\right) V=5 V $$
Since, the current is decreasing, the polarity of this emf would be so as to increase the existing current. The circuit can be redrawn as
Now,
$$ V _A-5+15+5=V _B $$
$$ \begin{array}{ll} \therefore & V _A-V _B=-15 V \\ \text { or } & V _B-V _A=15 V \end{array} $$
