A coil of inductance `0.20 H` is connected in series with a switch and a cell of emf `1.6 V`. The total resistance of the circuit is `4.0 Omega`. What is the initial rate of growth of the current when the switch is closed?

Options

A. `0.050 A s^(-1)`
B. `0.40 A s^(-1)`
C. `0.13 A s^(-1)`
D. `8.0 A s^(-1)`

Explanation

To find the initial rate of growth of the current when the switch is closed in an RL circuit, we can use the formula derived from Kirchhoff's voltage law. Here’s a step-by-step solution:

### Step 1: Write down the given values
- Inductance (L) = 0.20 H
- EMF (V) = 1.6 V
- Resistance (R) = 4.0 Ω

### Step 2: Use the formula for the series RL circuit
In a series RL circuit, the voltage equation is given by:
\[ V = R I + L \frac{dI}{dt} \]
Where:
- \( V \) is the EMF of the cell,
- \( R \) is the resistance,
- \( I \) is the current,
- \( L \) is the inductance,
- \( \frac{dI}{dt} \) is the rate of change of current.

### Step 3: Determine the initial conditions
At the moment the switch is closed (t = 0):
- The initial current \( I \) = 0 A (since the circuit has just been closed).

### Step 4: Substitute the initial conditions into the equation
At \( t = 0 \), the equation becomes:
\[ V = R(0) + L \frac{dI}{dt} \]
This simplifies to:
\[ V = L \frac{dI}{dt} \]

### Step 5: Solve for \( \frac{dI}{dt} \)
Rearranging the equation gives:
\[ \frac{dI}{dt} = \frac{V}{L} \]

### Step 6: Substitute the known values
Substituting the values of \( V \) and \( L \):
\[ \frac{dI}{dt} = \frac{1.6 \, \text{V}}{0.20 \, \text{H}} \]

### Step 7: Calculate \( \frac{dI}{dt} \)
\[ \frac{dI}{dt} = 8 \, \text{A/s} \]

### Conclusion
The initial rate of growth of the current when the switch is closed is \( 8 \, \text{A/s} \).

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