Electromagnetic Induction And Alternating Current Ques 44

Solution:

Formula:

Growth Of Current in Series R-L Circuit:

44. $\frac{d i}{d t}=10^{3} A / s$

$\therefore$ 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,

$$ V _A-5+15+5=V _B $$

$$ \begin{array}{ll} \therefore & V _A-V _B=-15 V \\ \text { or } & V _B-V _A=15 V \end{array} $$