An inductor L and a resistance R are connected in series with a battery of emf E and a switch. Initially the switch is open. The switch is closed at an instant t = 0. Select the correct alternatives

Options

A. At t = 0, induced emf in inductor is zero
B. At `t=(L)/(R)`, energy stored in the inductor is `(1-(1)/(e))^(2)((LE^(2))/(2R^(2)))`
C. At `t=(L)/(R)`, induced emf in the inductor is `(E)/(e)`
D. At `t=(L)/(R)`, voltage drop across the resistor is `E(1-(1)/(2))`

Answer

To solve the problem regarding the LR circuit connected to a battery, we will analyze the behavior of the circuit when the switch is closed at \( t = 0 \). We will derive the current, induced EMF, energy stored in the inductor, and the voltage drop across the resistor step by step.

### Step 1: Understanding the Circuit
When the switch is closed at \( t = 0 \), the circuit consists of an inductor \( L \), a resistor \( R \), and a battery of EMF \( E \). Initially, the current in the circuit is zero because the switch was open.

### Step 2: Current in the Circuit
The current \( I(t) \) in the circuit as a function of time \( t \) can be expressed using the formula:
\[ I(t) = \frac{E}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
At \( t = 0 \):
\[ I(0) = \frac{E}{R} \left(1 - e^{0}\right) = 0 \]
This means that initially, the current is zero.

### Step 3: Induced EMF in the Inductor
The induced EMF \( \mathcal{E} \) in the inductor can be calculated using:
\[ \mathcal{E} = -L \frac{dI}{dt} \]
Differentiating \( I(t) \):
\[ \frac{dI}{dt} = \frac{E}{R} \cdot \frac{R}{L} e^{-\frac{R}{L}t} = \frac{E}{L} e^{-\frac{R}{L}t} \]
Thus,
\[ \mathcal{E} = -L \cdot \frac{E}{L} e^{-\frac{R}{L}t} = -E e^{-\frac{R}{L}t} \]
At \( t = 0 \):
\[ \mathcal{E}(0) = -E \cdot e^{0} = -E \]
This indicates that the induced EMF is equal to \( -E \) at \( t = 0 \).

### Step 4: Energy Stored in the Inductor
The energy \( U \) stored in the inductor at time \( t \) is given by:
\[ U = \frac{1}{2} L I^2 \]
At \( t = \frac{L}{R} \):
\[ I\left(\frac{L}{R}\right) = \frac{E}{R} \left(1 - e^{-1}\right) \]
Substituting this into the energy formula:
\[ U = \frac{1}{2} L \left(\frac{E}{R} \left(1 - \frac{1}{e}\right)\right)^2 \]
Calculating this gives:
\[ U = \frac{L E^2}{2 R^2} \left(1 - \frac{1}{e}\right)^2 \]

### Step 5: Voltage Drop Across the Resistor
The voltage drop \( V_R \) across the resistor can be calculated using Ohm's law:
\[ V_R = I \cdot R \]
At \( t = \frac{L}{R} \):
\[ V_R = \left(\frac{E}{R} \left(1 - \frac{1}{e}\right)\right) R = E \left(1 - \frac{1}{e}\right) \]

### Summary of Results
1. At \( t = 0 \), the induced EMF in the inductor is \( -E \) (not zero).
2. At \( t = \frac{L}{R} \), the energy stored in the inductor is \( \frac{L E^2}{2 R^2} \left(1 - \frac{1}{e}\right)^2 \) (true statement).
3. At \( t = \frac{L}{R} \), the induced EMF is \( E/e \) (true statement).
4. The voltage drop across the resistor is \( E \left(1 - \frac{1}{e}\right) \) (this option is incorrect).