A coil of inductance 2H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, then the energy stored in the coil after 4s is J.

Answer

To find the energy stored in the coil after 4 seconds, we will follow these steps:

### Step 1: Understand the relationship between voltage, current, and inductance
The voltage \( V \) across an inductor is related to the rate of change of current \( I \) through it by the formula:
\[ V = L \frac{dI}{dt} \]
where \( L \) is the inductance of the coil.

### Step 2: Substitute the given voltage into the equation
Given that \( V = 3t \) volts and \( L = 2 \) H, we can write:
\[ 3t = 2 \frac{dI}{dt} \]

### Step 3: Rearrange the equation to find \( dI \)
Rearranging the equation gives:
\[ \frac{dI}{dt} = \frac{3t}{2} \]

### Step 4: Integrate to find the current \( I \)
Integrating both sides with respect to \( t \):
\[ I = \int \frac{3t}{2} dt = \frac{3}{2} \cdot \frac{t^2}{2} + C = \frac{3t^2}{4} + C \]
Since the current is zero at \( t = 0 \) (the coil is initially unenergized), we have \( C = 0 \). Thus:
\[ I = \frac{3t^2}{4} \]

### Step 5: Calculate the current at \( t = 4 \) seconds
Substituting \( t = 4 \) seconds into the current equation:
\[ I(4) = \frac{3(4^2)}{4} = \frac{3 \cdot 16}{4} = 12 \, \text{A} \]

### Step 6: Calculate the energy stored in the inductor
The energy \( W \) stored in an inductor is given by the formula:
\[ W = \frac{1}{2} L I^2 \]
Substituting \( L = 2 \, \text{H} \) and \( I = 12 \, \text{A} \):
\[ W = \frac{1}{2} \cdot 2 \cdot (12)^2 = 1 \cdot 144 = 144 \, \text{J} \]

### Final Answer
The energy stored in the coil after 4 seconds is \( 144 \, \text{J} \).

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