If a battery of emf `E` and internal resistance `r` is connected across a load of resistance `R`. Shot that the rate at which energy is dissipated in `R` is maximum when `R = r` and this maximur power is `P = E^2//4r`.

Explanation

To solve the problem, we need to show that the rate at which energy is dissipated in the load resistance \( R \) is maximum when \( R = r \) (where \( r \) is the internal resistance of the battery), and that this maximum power is given by \( P = \frac{E^2}{4r} \).

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a battery with an electromotive force (emf) \( E \) and an internal resistance \( r \).
- The battery is connected to an external load resistance \( R \).

2. **Finding the Current**:
- According to Kirchhoff's loop rule, the voltage across the load and the internal resistance must equal the emf of the battery:
\[ E - I(R + r) = 0 \]
- Rearranging gives us the expression for the current \( I \):
\[ I = \frac{E}{R + r} \]

3. **Calculating Power Dissipated in the Load**:
- The power \( P \) dissipated in the load resistance \( R \) can be expressed as:
\[ P = I^2 R \]
- 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. **Maximizing Power**:
- To find the maximum power, we need to differentiate \( P \) with respect to \( R \) and set the derivative to zero:
\[ \frac{dP}{dR} = \frac{d}{dR} \left(\frac{E^2 R}{(R + r)^2}\right) \]
- Using the quotient rule:
\[ \frac{dP}{dR} = \frac{(R + r)^2 \cdot E^2 - E^2 R \cdot 2(R + r)}{(R + r)^4} \]
- Setting the numerator equal to zero for maximization:
\[ (R + r)^2 E^2 - 2E^2 R(R + r) = 0 \]
- Simplifying:
\[ E^2 \left((R + r)^2 - 2R(R + r)\right) = 0 \]
- This leads to:
\[ (R + r)^2 - 2R(R + r) = 0 \]
- Expanding and simplifying:
\[ R^2 + 2Rr + r^2 - 2R^2 - 2Rr = 0 \implies -R^2 + r^2 = 0 \implies R = r \]

5. **Calculating Maximum Power**:
- Now substituting \( R = r \) back into the power equation:
\[ P = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{(2r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r} \]

### Final Result:
Thus, we have shown that the rate at which energy is dissipated in \( R \) is maximum when \( R = r \), and the maximum power is given by:
\[ P = \frac{E^2}{4r} \]