Current flowing through an inductor as a function of time is given as follows :
I = 4 + 16 t. here I is in amperes and t is in seconds. Emf induced in the inductor is 20 mV.
What is self-inductance of the inductor ?

Options

A. 1.25 `xx 10^(-3)` H
B. `2.5 xx 10^(-4)` H
C. `1.25 xx 10^(-4)` H
D. `5 xx 10^(-3)` H

Explanation

To find the self-inductance \( L \) of the inductor given the current \( I \) as a function of time and the induced EMF, we can follow these steps:

### Step 1: Identify the given information
We are given:
- The current \( I(t) = 4 + 16t \) (in amperes)
- The induced EMF \( E = 20 \, \text{mV} = 20 \times 10^{-3} \, \text{V} \)

### Step 2: Calculate the rate of change of current
To find the self-inductance, we need to calculate \( \frac{di}{dt} \). We differentiate the current function \( I(t) \):
\[ \frac{di}{dt} = \frac{d}{dt}(4 + 16t) = 16 \, \text{A/s} \]

### Step 3: Use the formula for induced EMF
The induced EMF in an inductor is given by the formula:
\[ E = -L \frac{di}{dt} \]
Since we are interested in the magnitude, we can drop the negative sign:
\[ E = L \frac{di}{dt} \]

### Step 4: Substitute the known values into the equation
Now we can substitute the values we have:
\[ 20 \times 10^{-3} = L \cdot 16 \]

### Step 5: Solve for self-inductance \( L \)
Rearranging the equation to solve for \( L \):
\[ L = \frac{20 \times 10^{-3}}{16} \]

Calculating this gives:
\[ L = 1.25 \times 10^{-3} \, \text{H} \]

### Step 6: Final answer
Thus, the self-inductance of the inductor is:
\[ L = 1.25 \, \text{mH} \text{ (milliHenries)} \]