Two identical batteries each of emf `E= 2` volt and internal resistance `r=1` ohm are available `t`. produce heat in an external resistance by passing a current through it. What is the maximum power that can be developed across an external resistance `R` using these batteries?

Explanation

To find the maximum power that can be developed across an external resistance \( R \) using two identical batteries, we will consider both series and parallel configurations of the batteries.

### Step 1: Analyze the Series Configuration
1. **Connect the batteries in series**: When connected in series, the total EMF \( E_{\text{total}} \) is the sum of the individual EMFs, and the total internal resistance \( r_{\text{total}} \) is the sum of the internal resistances.
\[ E_{\text{total}} = E + E = 2 + 2 = 4 \text{ volts} \]
\[ r_{\text{total}} = r + r = 1 + 1 = 2 \text{ ohms} \]

### Step 2: Calculate the Current in Series
2. **Use Ohm's Law to find the current \( I \)**:
The total resistance in the circuit is \( R + r_{\text{total}} \):
\[ I = \frac{E_{\text{total}}}{R + r_{\text{total}}} = \frac{4}{R + 2} \]

### Step 3: Calculate Power in Series
3. **Calculate the power \( P \) across the external resistance \( R \)**:
The power across the external resistance is given by:
\[ P = I^2 R = \left( \frac{4}{R + 2} \right)^2 R \]
To find the maximum power, we set \( R = r_{\text{total}} = 2 \) ohms:
\[ P_{\text{max}} = \left( \frac{4}{2 + 2} \right)^2 \cdot 2 = \left( \frac{4}{4} \right)^2 \cdot 2 = 1^2 \cdot 2 = 2 \text{ watts} \]

### Step 4: Analyze the Parallel Configuration
4. **Connect the batteries in parallel**: When connected in parallel, the EMF remains the same, but the internal resistance is halved:
\[ E_{\text{parallel}} = E = 2 \text{ volts} \]
\[ r_{\text{parallel}} = \frac{r}{2} = \frac{1}{2} \text{ ohm} \]

### Step 5: Calculate the Current in Parallel
5. **Use Ohm's Law to find the current \( I \)**:
The total resistance in the circuit is \( R + r_{\text{parallel}} \):
\[ I = \frac{E_{\text{parallel}}}{R + r_{\text{parallel}}} = \frac{2}{R + \frac{1}{2}} \]

### Step 6: Calculate Power in Parallel
6. **Calculate the power \( P \) across the external resistance \( R \)**:
The power across the external resistance is given by:
\[ P = I^2 R = \left( \frac{2}{R + \frac{1}{2}} \right)^2 R \]
To find the maximum power, we set \( R = r_{\text{parallel}} = \frac{1}{2} \) ohms:
\[ P_{\text{max}} = \left( \frac{2}{\frac{1}{2} + \frac{1}{2}} \right)^2 \cdot \frac{1}{2} = \left( \frac{2}{1} \right)^2 \cdot \frac{1}{2} = 4 \cdot \frac{1}{2} = 2 \text{ watts} \]

### Conclusion
In both configurations (series and parallel), the maximum power that can be developed across the external resistance \( R \) is:
\[ \text{Maximum Power} = 2 \text{ watts} \]