Question

The self-induced emf of a coil is $25\mathrm{volts}$. When the current in it is changed at a uniform rate from $10\mathrm{A}$to $25\mathrm{A}$ is $1\mathrm{s}$. What is the change in the energy of the inductance?

• A. $637.5\text{J}$
• B. $437.5\mathrm{J}$
• C. $540\mathrm{J}$
• D. $740\mathrm{J}$

Answer 1

The correct option is B

$437.5\mathrm{J}$

Step 1: Given data

Initial Current, $i_1=10\mathrm{A}$

Final Current, $i_2=25\mathrm{A}$

$\mathrm{emf}=25\mathrm{V}$

$dt=1\text{s}$

Step 2: Formula used

$\mathrm{emf}=L\frac{di}{dt}$,

Where $L$is Inductance, $\frac{di}{dt}$ is the rate of change of current with time.

$E=\frac{1}{2}Li$

Where,

$E$ is the energy stored.

$L$ is the inductance.

$i$ is the current.

Step 3: Calculating the change in stored energy

A magnetic field is created inside a coil as a current passes through it, as we know. The flux through the coil fluctuates when the current is constant and changes over time. An emf is induced in the coil due to a change in flux through the coil.

The magnitude of induced $emf=L\frac{di}{dt}$ -------- (i)

The emf induced resists change in current. To put it another way, it serves as an opposing force, storing energy in the form of a magnetic field inside the coil. If the coil's current is $i$, the stored energy is given by,

$E=\frac{1}{2}Li$ ------- (ii)

When the current changes from $i_1$ to $i_2$, then the change in energy of inductance will be

$$\begin{gathered} ∆E=E_2-E_1 \\ ∆E=\frac{1}{2}L{\left(i_2\right)}^2-\frac{1}{2}L{\left(i_1\right)}^2 \\ ∆E=\frac{1}{2}L\left[{\left(i_2\right)}^2-{\left(i_1\right)}^2\right] \end{gathered}$$

Value of $i_1=10A$ and $i_2=25A$

$∆E=\frac{1}{2}L({25}^2-{10}^2)$

$$\begin{gathered} ∆E=\frac{1}{2}L(625-100) \\ ∆E=\frac{1}{2}L\left(525\right) \end{gathered}$$

$∆E=232.5·L$ -------- (iii)

Step 4: Calculating inductance

The current changes at a constant rate, going from $10\mathrm{A}$to$25\mathrm{A}$in $1\mathrm{s}$. This means,

$\frac{di}{dt}=\frac{∆i}{∆t}=\frac{25-10}{1}=15A{s}^{-1}$

Substitute the values of $emf$ and $\frac{di}{dt}$ in equation (i).

$$\begin{gathered} \Rightarrow 25=15L \\ \Rightarrow L=\frac{25}{15}=\frac{5}{3}H \end{gathered}$$

Step 5: Calculating change in stored energy of the inductor

Substitute value of L in equation (iii)

$∆E=232.5\times \frac{5}{3}=437.5\mathrm{J}$

Hence, option (B) is correct, that is $437.5\mathrm{J}$.