A coil of inductance 0.2 H and 1.0 ohm resistance is connected to a 90 V source. At what rate will the current in the coil grow at the instant the coil is connected to the source?

Options

A. `450 As^(-1)`
B. `4.5As^(-1)`
C. `45As^(-1)`
D. `0.45As^(-1)`

Answer

To solve the problem, we need to determine the rate of change of current (\( \frac{dI}{dt} \)) in an RL circuit at the moment the coil is connected to a voltage source.

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance (\( L \)) = 0.2 H
- Resistance (\( R \)) = 1.0 Ω
- Voltage (\( E \)) = 90 V

2. **Use the formula for the rate of change of current in an RL circuit:**
The formula relating the voltage, inductance, and rate of change of current is given by:
\[ E = L \frac{dI}{dt} + I R \]
At the instant the coil is connected to the source, the current (\( I \)) is 0 (since it has just been connected). Thus, the equation simplifies to:
\[ E = L \frac{dI}{dt} \]

3. **Rearranging the equation to find \( \frac{dI}{dt} \):**
\[ \frac{dI}{dt} = \frac{E}{L} \]

4. **Substituting the known values into the equation:**
\[ \frac{dI}{dt} = \frac{90 \text{ V}}{0.2 \text{ H}} = 450 \text{ A/s} \]

5. **Conclusion:**
The rate at which the current in the coil grows at the instant the coil is connected to the source is \( 450 \text{ A/s} \).

### Final Answer:
The rate of change of current in the coil at the instant it is connected to the source is **450 A/s**.

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