A circuit consists of a coil, a switch and a battery, all in series. The internal resistance of the battery is negligible compared with that of the coil. The switch is originally open. It is thrown closed, after a time interval `Deltat`. The current in the circuit reaches `80.0%` of its final value. The switch then remains closed for a time interval much longer than `Deltat`. The wires connected to the terminals of the battery are then short circuited with another wire and removed from the battery, so that the current is uninterrupted.
At an instant that is a time interval `Deltat`. After the short circuit, the current is what percentage of its maximum value? a)20% b) 8% c) 4% d)10 %

Options

A. `20.0%`
B. `8.0%`
C. `4.0%`
D. `10.0%`

Answer

To solve the problem step by step, we will analyze the circuit behavior before and after the switch is short-circuited.

### Step 1: Understanding the Circuit
The circuit consists of a coil (inductor), a switch, and a battery in series. The internal resistance of the battery is negligible compared to the resistance of the coil.

### Step 2: Current Growth in the Inductor
When the switch is closed, the current in the inductor grows according to the formula:
\[ I(t) = \frac{E}{R} \left(1 - e^{-\frac{L t}{R}}\right) \]
where:
- \( E \) is the EMF of the battery,
- \( R \) is the resistance of the coil,
- \( L \) is the inductance of the coil,
- \( t \) is the time after the switch is closed.

### Step 3: Current After Time Interval \( \Delta t \)
After a time interval \( \Delta t \), the problem states that the current reaches 80% of its maximum value:
\[ I(\Delta t) = 0.8 I_{\text{max}} \]
where \( I_{\text{max}} = \frac{E}{R} \).

Substituting this into the current equation:
\[ 0.8 \frac{E}{R} = \frac{E}{R} \left(1 - e^{-\frac{L \Delta t}{R}}\right) \]

### Step 4: Simplifying the Equation
Dividing both sides by \( \frac{E}{R} \):
\[ 0.8 = 1 - e^{-\frac{L \Delta t}{R}} \]
Rearranging gives:
\[ e^{-\frac{L \Delta t}{R}} = 1 - 0.8 = 0.2 \]

### Step 5: Finding \( \Delta t \)
Taking the natural logarithm of both sides:
\[ -\frac{L \Delta t}{R} = \ln(0.2) \]
Thus:
\[ \Delta t = -\frac{R}{L} \ln(0.2) \]

### Step 6: Current After Short Circuit
After the switch is closed for a long time, the current reaches its maximum value \( I_{\text{max}} \). When the battery is short-circuited, the current starts to decay according to the formula:
\[ I(t) = I_{\text{max}} e^{-\frac{R t}{L}} \]

### Step 7: Current After Time Interval \( \Delta t \) Post Short Circuit
We need to find the current \( I \) at a time interval \( \Delta t \) after the short circuit:
\[ I(\Delta t) = I_{\text{max}} e^{-\frac{R \Delta t}{L}} \]

Substituting \( \Delta t = -\frac{R}{L} \ln(0.2) \):
\[ I(\Delta t) = I_{\text{max}} e^{-\frac{R}{L} \cdot -\frac{R}{L} \ln(0.2)} \]
This simplifies to:
\[ I(\Delta t) = I_{\text{max}} e^{\ln(0.2)} \]
\[ I(\Delta t) = I_{\text{max}} \cdot 0.2 \]

### Step 8: Finding the Percentage of Maximum Current
Thus, the current \( I \) at time \( \Delta t \) after the short circuit is:
\[ I = 0.2 I_{\text{max}} \]
This means the current is 20% of its maximum value.

### Final Answer
The correct answer is:
**a) 20%**