A cell of internal resistance 2Ω supplies current to a 6Ω resistor. The efficiency of the cell is

1. Analyze the question

We are given a cell with an internal resistance ($r$) of $2\Omega$ connected to an external load resistor ($R$) of $6\Omega$. We need to calculate the efficiency of the cell.

2. Explain the concept

The efficiency ($\eta$) of a cell is defined as the ratio of the useful power delivered to the external resistor to the total power generated by the cell.

Mathematically, it is expressed as:

Since Power $P = I^2R$, we can write:

By cancelling out the $I^2$ (current squared), the formula simplifies to:

3. Show the solution

• Step 1: Identify the given values • External resistance ($R$) = $6\Omega$ • Internal resistance ($r$) = $2\Omega$

• Step 2: Substitute the values into the formula $$\eta = \frac{6}{6 + 2} \times 100\%$$

• Step 3: Perform the calculation $$\eta = \frac{6}{8} \times 100\%$$ $$\eta = 0.75 \times 100\%$$ $$\eta = 75.0\%$$

4. Identify the answer

Comparing our result to the given options, the correct choice is D.

5. Exam Tip

Whenever you see a question about the efficiency of a cell, remember the "Voltage Ratio" trick: Efficiency is simply the ratio of the Terminal Voltage ($V$) to the Electromotive Force ($E$). Since $V = IR$ and $E = I(R+r)$, the current $I$ always cancels out, leaving you with just the resistances. Don't waste time trying to find the current!