Electromagnetic Induction And Alternating Current Ques 16
- The self-induced emf of a coil is $25 V$. When the current in it is changed at uniform rate from $10 A$ to $25 A$ in $1 s$, the change in the energy of the inductance is
(Main 2019, 10 Jan II)
(a) $437.5 J$
(b) $740 J$
(c) $637.5 J$
(d) $540 J$
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Answer:
Correct Answer: 16.(a)
Solution:
- Energy stored in an inductor of inductance $L$ and current $I$ is given by
$$ E=\frac{1}{2} L I^{2} $$
When current is being changed from $I _1$ to $I _2$, change in energy will be
$$ \Delta E=E _2-E _1=\frac{1}{2} L I _2^{2}-\frac{1}{2} L I _1^{2} $$
As $I _1$ and $I _2$ are given, we need to find value of $L$.
Now, induced emf in a coil is
$$ \varepsilon=L \frac{d I}{d t} $$
Here,
$$ \varepsilon=25 V $$
$$ d I=I _2-I _1=(25-10)=15 A \text { and } d t=1 s $$
$$ \Rightarrow \quad 25=L \times \frac{15}{1} \text { or } L=\frac{25}{15}=5 / 3 H $$
Putting values of $L, I _1$ and $I _2$ in Eq. (i), we get
$$ \Delta E=\frac{1}{2} \times \frac{5}{3} \times\left[25^{2}-10^{2}\right]=\frac{1}{2} \times \frac{5}{3} \times 525 \Delta E=437.5 J $$
BETA
