Rate of increment of energy in an inductor with time in series RL circuit getting charged with battery of EMF Eis best represented by:

Options

A.
B.
C.
D.

Answer

To solve the problem of finding the rate of increment of energy in an inductor with time in a series RL circuit being charged by a battery of EMF \( E \), we can follow these steps:

### Step 1: Understand the Energy in an Inductor
The energy stored in an inductor \( U \) is given by the formula:
\[ U = \frac{1}{2} L I^2 \]
where \( L \) is the inductance and \( I \) is the current through the inductor.

### Step 2: Differentiate the Energy with Respect to Time
To find the rate of change of energy with respect to time, we differentiate \( U \):
\[ \frac{dU}{dt} = \frac{d}{dt} \left( \frac{1}{2} L I^2 \right) \]
Since \( L \) is a constant, we can take it out of the differentiation:
\[ \frac{dU}{dt} = L \cdot I \cdot \frac{dI}{dt} \]
This is our Equation (1).

### Step 3: Determine the Current in the RL Circuit
The current \( I \) in an RL circuit at any time \( t \) can be expressed as:
\[ I = I_0 \left( 1 - e^{-\frac{t}{\tau}} \right) \]
where \( I_0 \) is the maximum current and \( \tau = \frac{L}{R} \) is the time constant of the circuit.

### Step 4: Differentiate the Current with Respect to Time
Now, we differentiate \( I \) with respect to time \( t \):
\[ \frac{dI}{dt} = I_0 \cdot \frac{d}{dt} \left( 1 - e^{-\frac{t}{\tau}} \right) \]
The derivative of \( e^{-\frac{t}{\tau}} \) is:
\[ -\frac{1}{\tau} e^{-\frac{t}{\tau}} \]
Thus,
\[ \frac{dI}{dt} = I_0 \cdot \left( 0 + \frac{1}{\tau} e^{-\frac{t}{\tau}} \right) = \frac{I_0}{\tau} e^{-\frac{t}{\tau}} \]
This is our Equation (2).

### Step 5: Substitute \( \frac{dI}{dt} \) into \( \frac{dU}{dt} \)
Now we substitute Equation (2) into Equation (1):
\[ \frac{dU}{dt} = L \cdot I \cdot \frac{dI}{dt} = L \cdot I_0 \left( 1 - e^{-\frac{t}{\tau}} \right) \cdot \frac{I_0}{\tau} e^{-\frac{t}{\tau}} \]
This simplifies to:
\[ \frac{dU}{dt} = \frac{L I_0^2}{\tau} \left( 1 - e^{-\frac{t}{\tau}} \right) e^{-\frac{t}{\tau}} \]

### Step 6: Analyze the Behavior at \( t = 0 \) and \( t \to \infty \)
- At \( t = 0 \):
\[ \frac{dU}{dt} = \frac{L I_0^2}{\tau} \cdot 0 \cdot 1 = 0 \]
- As \( t \to \infty \):
\[ \frac{dU}{dt} = \frac{L I_0^2}{\tau} \cdot 1 \cdot 0 = 0 \]

### Conclusion
The rate of increment of energy in the inductor starts from zero, increases to a maximum, and then returns to zero as the current stabilizes. Therefore, the best representation of the rate of increment of energy in the inductor over time is a function that starts and ends at zero.

### Final Answer
The correct option is **A**.