A cell of internal resistances `r` is connected to a load of resistance `R` . Energy is dissipated in the load, but some thermal energy is also wasted in the cell. The efficiency of such an arrangement is found from the expression
( Energy dissipated in the load )/( Energy dissipated in the complete circuit)
Which of the following gives the efficiency in this case?

Options

A. `( r ) /( R )`
B. `( R )/( r)`
C. `( r )/( R + r)`
D. `( R ) /( R + r )`

Explanation

To find the efficiency of a cell connected to a load resistance, we can follow these steps:

### Step 1: Understand the Circuit
We have a cell with an internal resistance \( r \) connected to a load resistance \( R \). The current \( I \) flowing through the circuit can be calculated using Ohm's law.

### Step 2: Calculate the Current
The total resistance in the circuit is the sum of the internal resistance and the load resistance:
\[ R_{total} = R + r \]
Using Ohm's law, the current \( I \) flowing through the circuit is given by:
\[ I = \frac{E}{R + r} \]
where \( E \) is the electromotive force (emf) of the cell.

### Step 3: Calculate the Heat Dissipated in the Load
The heat \( H_R \) dissipated in the load resistance \( R \) can be calculated using the formula:
\[ H_R = I^2 R \]
Substituting the expression for current:
\[ H_R = \left(\frac{E}{R + r}\right)^2 R = \frac{E^2 R}{(R + r)^2} \]

### Step 4: Calculate the Total Heat Produced
The total heat \( H_{total} \) produced in the circuit includes the heat dissipated in both the load and the internal resistance:
\[ H_{total} = H_R + H_r \]
where \( H_r \) is the heat dissipated in the internal resistance \( r \):
\[ H_r = I^2 r = \left(\frac{E}{R + r}\right)^2 r = \frac{E^2 r}{(R + r)^2} \]
Thus,
\[ H_{total} = \frac{E^2 R}{(R + r)^2} + \frac{E^2 r}{(R + r)^2} = \frac{E^2 (R + r)}{(R + r)^2} \]

### Step 5: Calculate the Efficiency
Efficiency \( \eta \) is defined as the ratio of the energy dissipated in the load to the total energy dissipated in the circuit:
\[ \eta = \frac{H_R}{H_{total}} = \frac{\frac{E^2 R}{(R + r)^2}}{\frac{E^2 (R + r)}{(R + r)^2}} = \frac{R}{R + r} \]

### Conclusion
The efficiency of the arrangement is given by:
\[ \eta = \frac{R}{R + r} \]