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. `0.5 Omega`
B. `1.0 Omega`
C. `2.0 Omega`
D. `0.25 Omega`

Answer

To find the value of the variable resistance \( R \) for which the power delivered in it is maximum, we can follow these steps:

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

### Step 2: Write the Expression for Power
The total resistance in the circuit is the sum of the internal resistance and the variable resistance:
\[ R_{\text{total}} = R + r = R + 0.5 \, \Omega \]
The power \( P \) delivered to the resistor \( R \) can be expressed as:
\[ P = \frac{V^2}{R_{\text{total}}^2} \cdot R = \frac{10^2}{(R + 0.5)^2} \cdot R \]
This simplifies to:
\[ P = \frac{100R}{(R + 0.5)^2} \]

### Step 3: Differentiate Power with Respect to R
To find the maximum power, we need to differentiate \( P \) with respect to \( R \) and set the derivative equal to zero:
\[ \frac{dP}{dR} = \frac{(R + 0.5)^2 \cdot 100 - 100R \cdot 2(R + 0.5)}{(R + 0.5)^4} \]
Setting the numerator equal to zero gives:
\[ (R + 0.5)^2 \cdot 100 - 200R(R + 0.5) = 0 \]

### Step 4: Solve the Equation
Expanding and simplifying the equation:
\[ 100(R + 0.5)^2 = 200R(R + 0.5) \]
\[ 100(R^2 + R + 0.25) = 200R^2 + 100R \]
\[ 100R^2 + 100R + 25 = 200R^2 + 100R \]
\[ 0 = 100R^2 - 25 \]
\[ 100R^2 = 25 \]
\[ R^2 = 0.25 \]
\[ R = 0.5 \, \Omega \]

### Step 5: Conclusion
The value of \( R \) for which the power delivered in it is maximum is:
\[ \boxed{0.5 \, \Omega} \]

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