A capacitor of capacitance `C` is charge by a battery of emf `E` and internal resistance r. A resistasnce 2r is also connet in sereis with the capacitor. The amount of heat liberated inside the battery by the time capacitor is `50%` charged is

Options

A. `3/8E^2C`
B. `(E^2C)/6`
C. `(E^2C)/12`
D. `(E^2C)/24`

Explanation

To solve the problem, we need to determine the amount of heat liberated inside the battery by the time the capacitor is 50% charged. Here’s a step-by-step solution:

### Step 1: Understand the Energy Supplied by the Battery
When a capacitor is charged, the energy supplied by the battery is partly stored in the capacitor and partly lost as heat. The total energy \( U \) supplied by the battery when the capacitor is fully charged is given by:
\[ U = \frac{1}{2} C V^2 \]
where \( V \) is the voltage across the capacitor.

### Step 2: Determine the Voltage Across the Capacitor
In this case, the capacitor is charged by a battery with an emf \( E \) and an internal resistance \( r \). Additionally, there is a resistance \( 2r \) in series. The total resistance in the circuit is \( R_{total} = r + 2r = 3r \).

### Step 3: Calculate the Current in the Circuit
The current \( I \) in the circuit can be calculated using Ohm's law:
\[ I = \frac{E}{R_{total}} = \frac{E}{3r} \]

### Step 4: Determine the Charge on the Capacitor
The charge \( Q \) on the capacitor after a time \( t \) can be expressed as:
\[ Q = C \cdot V = C \cdot \left( E \left(1 - e^{-\frac{t}{RC}}\right) \right) \]
However, since we need the heat loss when the capacitor is 50% charged, we can simplify our calculations.

### Step 5: Calculate the Energy Lost During Charging
When the capacitor is 50% charged, the voltage across it is \( \frac{E}{2} \). The energy stored in the capacitor at this point is:
\[ U_{stored} = \frac{1}{2} C \left(\frac{E}{2}\right)^2 = \frac{1}{2} C \cdot \frac{E^2}{4} = \frac{E^2 C}{8} \]

### Step 6: Calculate the Total Energy Supplied by the Battery
The total energy supplied by the battery until the capacitor is 50% charged can be calculated by considering the energy lost:
\[ U_{total} = U_{stored} + U_{lost} \]
Since 50% of the energy is lost, we have:
\[ U_{total} = U_{stored} + U_{stored} = 2 \cdot U_{stored} = 2 \cdot \frac{E^2 C}{8} = \frac{E^2 C}{4} \]

### Step 7: Calculate the Heat Lost in the Battery
The heat lost in the battery can be calculated as a fraction of the total energy lost. The ratio of the resistances is \( r : 2r = 1 : 2 \), so the heat loss in the battery is:
\[ U_{battery} = \frac{1}{3} U_{lost} \]
Substituting \( U_{lost} = \frac{E^2 C}{8} \):
\[ U_{battery} = \frac{1}{3} \cdot \frac{E^2 C}{8} = \frac{E^2 C}{24} \]

### Final Answer
Thus, the amount of heat liberated inside the battery by the time the capacitor is 50% charged is:
\[ \boxed{\frac{E^2 C}{24}} \]