Question

Self inductance of a coil is $8H$ . The power (in watt) consumed by coil (primary inductive) is given by $P = 8{i^2}$ where $'i'$ current in ampere. The time for the current to change from ${i_0}$ to $2{i_0}$ will be
(A) $\ln 2$
(B) $5{i_0}$
(C) $\dfrac{{{i_0}^2}}{{\ln 2}}$
(D) None of these

Answer 1

To solve this question, we need to use the formula for the energy stored by an inductor and differentiate it to get the expression for the power. On equating it with the expression for the power given in the question, we will get a differential equation in the current. On integrating the differential equation between the given limits, we will get the value of the time.

Formula used: The formula used to solve this question is given by
$E = \dfrac{1}{2}L{i^2}$ , here $E$ is the energy stored by an inductor of inductance $L$ through which a current of $i$ is flowing.

Complete Step-by-Step solution:
Let $T$ be the time for the given change in the current.
We know that the energy stored in an inductor as a function of the current flowing through it is given by
$E = \dfrac{1}{2}L{i^2}$ ......................(1)
Now, we know that the power is equal to the rate of energy supplied. Mathematically, this is given by
$P = \dfrac{{dE}}{{dt}}$ ......................(2)
Therefore differentiating both sides of (1) we get
$\dfrac{{dE}}{{dt}} = \dfrac{{d\left( {\dfrac{1}{2}L{i^2}} \right)}}{{dt}}$
$\Rightarrow \dfrac{{dE}}{{dt}} = \dfrac{1}{2}L \times 2i\dfrac{{di}}{{dt}}$
On simplifying we get
$\dfrac{{dE}}{{dt}} = Li\dfrac{{di}}{{dt}}$ ......................(3)
From (2) and (3) we can write
$Li\dfrac{{di}}{{dt}} = P$ ......................(4)
According to the question, we have
$L = 8H$ ......................(5)
$P = 8{i^2}$ ......................(6)
Putting (5) and (6) in (4) we get
$8i\dfrac{{di}}{{dt}} = 8{i^2}$
Dividing both the sides by $8i$ we have
$\dfrac{{di}}{{dt}} = i$
On rearranging the terms, we can write the above equation as
$dt = \dfrac{{di}}{i}$
Integrating both the sides, we get
$\int\limits_0^T {dt} = \int\limits_{{i_0}}^{2{i_0}} {\dfrac{{di}}{i}}$
$\Rightarrow \left[ t \right]_0^T = \left[ {\ln i} \right]_{{i_0}}^{2{i_0}}$
Substituting the limits, we get
$T - 0 = \ln 2{i_0} - \ln {i_0}$
$\Rightarrow T = \ln 2{i_0} - \ln {i_0}$
We know that $\ln A - \ln B = \ln \dfrac{A}{B}$ . So the above expression for the time can be written as
$T = \ln \left( {\dfrac{{2{i_0}}}{{{i_0}}}} \right)$
On simplifying, we finally get
$T = \ln 2$
Thus, the time required for the current to change from ${i_0}$ to $2{i_0}$ is equal to $\ln 2$ .
Hence, the correct answer is option (A).

Note:
Do not consider the power and the energy to be the same. We must note that we have been given the expression for the power consumed by the coil, and not the energy.