A cell of internal resistance r drivers current through an external resistance R. The power delivered by the to the external resistance will be maximum when:

Options

A. `R=0.001r`
B. `R=1000r`
C. `R=2r`
D. `R=r`

Answer

To solve the problem of finding the condition for maximum power delivered to an external resistance \( R \) by a cell with internal resistance \( r \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circuit**: We have a cell with an electromotive force (emf) \( E \) and an internal resistance \( r \). This cell is connected to an external resistance \( R \). The total resistance in the circuit is \( R + r \).

2. **Calculate the Current**: The current \( I \) flowing through the circuit can be calculated using Ohm's Law:
\[ I = \frac{E}{R + r} \]

3. **Calculate the Power Delivered to the External Resistance**: The power \( P \) delivered to the external resistance \( R \) is given by:
\[ P = I^2 R \]
Substituting the expression for \( I \):
\[ P = \left(\frac{E}{R + r}\right)^2 R \]
Simplifying this, we get:
\[ P = \frac{E^2 R}{(R + r)^2} \]

4. **Maximize the Power**: To find the maximum power, we need to differentiate \( P \) with respect to \( R \) and set the derivative equal to zero:
\[ \frac{dP}{dR} = 0 \]

5. **Differentiate the Power Function**: Using the quotient rule, we differentiate \( P \):
\[ \frac{dP}{dR} = \frac{(R + r)^2 \cdot E^2 - E^2 R \cdot 2(R + r)}{(R + r)^4} \]
Setting the numerator equal to zero for maximization:
\[ (R + r)^2 - 2R(R + r) = 0 \]

6. **Simplify the Equation**: Expanding and simplifying the equation:
\[ R^2 + 2Rr + r^2 - 2R^2 - 2Rr = 0 \]
\[ -R^2 + r^2 = 0 \]
\[ r^2 = R^2 \]
This implies:
\[ r = R \]

7. **Conclusion**: The power delivered to the external resistance is maximum when the internal resistance \( r \) is equal to the external resistance \( R \).

### Final Answer:
The power delivered by the cell to the external resistance will be maximum when \( r = R \).

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