A battery of e.m.f. `10 V` and internal resistance `0.5 ohm` is connected across a variable resistance `R`. The value of `R` for which the power delivered in it is maximum is given by

Options

A. `2.0 ohm`
B. `0.25 ohm`
C. `1.0 ohm`
D. `0.5 ohm`

Explanation

To solve the problem, we need to determine the value of the variable resistance \( R \) for which the power delivered in it is maximum when connected to a battery with an e.m.f. of \( 10 \, \text{V} \) and an internal resistance of \( 0.5 \, \Omega \).

### Step-by-Step Solution:

1. **Understand the Circuit**:
We have a battery with an e.m.f. \( E = 10 \, \text{V} \) and an internal resistance \( r = 0.5 \, \Omega \). The battery is connected to a variable external resistance \( R \).

2. **Power Delivered to the Load**:
The power \( P \) delivered to the external resistance \( R \) can be expressed using the formula:
\[ P = I^2 R \]
where \( I \) is the current flowing through the circuit.

3. **Calculate the Current**:
The total resistance in the circuit is \( R + r \). According to Ohm's law, the current \( I \) can be calculated as:
\[ I = \frac{E}{R + r} = \frac{10}{R + 0.5} \]

4. **Substituting Current into Power Formula**:
Now, substituting \( I \) into the power equation gives:
\[ P = \left(\frac{10}{R + 0.5}\right)^2 R \]
Simplifying this, we get:
\[ P = \frac{100 R}{(R + 0.5)^2} \]

5. **Finding Maximum Power**:
To find the value of \( R \) that maximizes power \( P \), we can use the condition from the maximum power transfer theorem, which states that maximum power is delivered when the internal resistance \( r \) is equal to the external resistance \( R \):
\[ R = r \]

6. **Substituting the Internal Resistance**:
Given that \( r = 0.5 \, \Omega \), we find:
\[ R = 0.5 \, \Omega \]

### Final Answer:
The value of \( R \) for which the power delivered in it is maximum is \( 0.5 \, \Omega \).

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