A coil of resistance `10Omega` and an inductance `5H` is connected to a `100` volt battery. Then energy stored in the coil is

Options

A. 125 erg
B. 125 J
C. 250 erg
D. 250 J

Answer

To find the energy stored in the coil, we can follow these steps:

### Step 1: Determine the steady-state current through the coil
In steady state, the inductor behaves like a short circuit (or a plain wire). The current \( I \) through the coil can be calculated using Ohm's Law:

\[ I = \frac{V}{R} \]

Where:
- \( V = 100 \, \text{volts} \) (voltage of the battery)
- \( R = 10 \, \Omega \) (resistance of the coil)

Substituting the values:

\[ I = \frac{100 \, \text{V}}{10 \, \Omega} = 10 \, \text{A} \]

### Step 2: Calculate the energy stored in the inductor
The energy \( U \) stored in an inductor can be calculated using the formula:

\[ U = \frac{1}{2} L I^2 \]

Where:
- \( L = 5 \, \text{H} \) (inductance of the coil)
- \( I = 10 \, \text{A} \) (current calculated in Step 1)

Substituting the values:

\[ U = \frac{1}{2} \times 5 \, \text{H} \times (10 \, \text{A})^2 \]
\[ U = \frac{1}{2} \times 5 \times 100 = \frac{500}{2} = 250 \, \text{J} \]

### Final Answer
The energy stored in the coil is \( 250 \, \text{J} \).

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