The current passing through a choke coil of `5` henry is decreasing at the rate of `2 ampere//sec`. The e.mf. Developing across the coil is

Options

A. 10 V
B. `-10 V`
C. 2.5 V
D. `-2.5 mV`

Explanation

To find the electromotive force (e.m.f.) developed across a choke coil, we can use the formula related to self-inductance. Here’s a step-by-step solution:

### Step 1: Understand the given values
- Inductance (L) of the choke coil = 5 Henry
- Rate of change of current (di/dt) = -2 A/s (the negative sign indicates that the current is decreasing)

### Step 2: Use the formula for e.m.f.
The formula for the e.m.f. (ε) induced in an inductor due to a change in current is given by:
\[ \epsilon = -L \frac{di}{dt} \]
Where:
- ε is the induced e.m.f.
- L is the inductance
- di/dt is the rate of change of current

### Step 3: Substitute the values into the formula
Substituting the values we have:
\[ \epsilon = -5 \, \text{H} \times \left(-2 \, \frac{\text{A}}{\text{s}}\right) \]

### Step 4: Calculate the e.m.f.
Now, calculate the e.m.f.:
\[ \epsilon = -5 \times -2 = 10 \, \text{V} \]

### Step 5: State the final answer
The e.m.f. developed across the coil is:
\[ \epsilon = 10 \, \text{V} \]