An total resistance `R` is connected to a cell of internal resistance `r` the maximum current flows in the external resistance, when

Options

A. `R = r`
B. `R lt r`
C. `R gt r`
D. none of the above

Answer

To solve the problem, we need to determine the condition under which the maximum current flows through the external resistance \( R \) connected to a cell with internal resistance \( r \).

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a cell with an electromotive force (emf) \( e \) and an internal resistance \( r \).
- The external resistance is \( R \).
- The total resistance in the circuit is \( R + r \).

2. **Applying Ohm's Law**:
- The current \( I \) flowing through the circuit can be expressed using Ohm's law:
\[ I = \frac{e}{R + r} \]

3. **Power Dissipated in the External Resistance**:
- The power \( P \) dissipated in the external resistance \( R \) is given by:
\[ P = I^2 R = \left(\frac{e}{R + r}\right)^2 R \]
- Simplifying this, we get:
\[ P = \frac{e^2 R}{(R + r)^2} \]

4. **Finding Maximum Power**:
- To find the condition for maximum power, we need to differentiate \( P \) with respect to \( R \) and set the derivative equal to zero:
\[ \frac{dP}{dR} = \frac{d}{dR} \left(\frac{e^2 R}{(R + r)^2}\right) \]
- Using the quotient rule for differentiation:
\[ \frac{dP}{dR} = \frac{(R + r)^2 \cdot e^2 - e^2 R \cdot 2(R + r)}{(R + r)^4} \]
- Setting the numerator equal to zero for maximum power:
\[ (R + r)^2 e^2 - 2 e^2 R (R + r) = 0 \]
- Simplifying this leads to:
\[ e^2 (R + r - 2R) = 0 \]
\[ R + r - 2R = 0 \implies r = R \]

5. **Conclusion**:
- The maximum current flows through the external resistance \( R \) when the external resistance \( R \) is equal to the internal resistance \( r \).

### Final Answer:
The maximum current flows in the external resistance when \( R = r \).