To solve the problem, we need to find the internal resistance \( r \) of a cell that supplies current to a 6.0Ω resistor with an efficiency of 75%. Let's break down the steps to arrive at the correct answer. ### Step 1: Understanding Efficiency Efficiency (\( \eta \)) is defined as the ratio of useful power output to the total power input, expressed as a percentage. In this case, the useful power output is the power delivered to the external resistor (6.0Ω), and the total power input is the power supplied by the cell. The formula for efficiency is given by: \[ \eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% \] Where: - \( P_{\text{out}} \) is the power across the external resistor. - \( P_{\text{in}} \) is the total power supplied by the cell. ### Step 2: Power Calculations The power across the external resistor can be calculated using Ohm's law. The current \( I \) flowing through the resistor can be expressed in terms of the voltage \( V \) of the cell and the total resistance in the circuit, which includes both the external resistor and the internal resistance of the cell. The total resistance \( R_{\text{total}} \) in the circuit is: \[ R_{\text{total}} = R + r = 6.0Ω + r \] Using Ohm's law, the current \( I \) can be expressed as: \[ I = \frac{V}{R_{\text{total}}} = \frac{V}{6.0 + r} \] The power across the external resistor is: \[ P_{\text{out}} = I^2 R = \left(\frac{V}{6.0 + r}\right)^2 \cdot 6.0 \] The total power supplied by the cell is: \[ P_{\text{in}} = I \cdot V = \frac{V^2}{6.0 + r} \] ### Step 3: Setting Up the Efficiency Equation Now we can substitute \( P_{\text{out}} \) and \( P_{\text{in}} \) into the efficiency formula: \[ 75\% = \frac{\left(\frac{V}{6.0 + r}\right)^2 \cdot 6.0}{\frac{V^2}{6.0 + r}} \times 100\% \] ### Step 4: Simplifying the Equation We can simplify the equation: \[ 0.75 = \frac{\left(\frac{V^2 \cdot 6.0}{(6.0 + r)^2}\right)}{\frac{V^2}{6.0 + r}} \] Cancelling \( V^2 \) from both sides gives: \[ 0.75 = \frac{6.0}{6.0 + r} \] ### Step 5: Solving for \( r \) Now we can solve for \( r \): \[ 0.75(6.0 + r) = 6.0 \] \[ 4.5 + 0.75r = 6.0 \] \[ 0.75r = 6.0 - 4.5 \] \[ 0.75r = 1.5 \] \[ r = \frac{1.5}{0.75} = 2.0Ω \] ### Conclusion The internal resistance \( r \) of the cell is **2.0Ω**. Therefore, the correct option is **D**. ### Explanation of Other Options - **A. 4.5Ω**: This value does not satisfy the efficiency condition derived from the power equations. - **B. 1.0Ω**: This would result in a higher efficiency than 75%, which contradicts the given efficiency. - **C. 8.0Ω**: This would lead to a much lower efficiency than 75%, as the internal resistance would dominate the total resistance. ### Revision Summary - Efficiency is the ratio of useful power output to total power input. - Use Ohm's law to relate current, voltage, and resistance in power calculations. - Set up the efficiency equation and simplify to find the internal resistance. - The internal resistance of the cell in this problem is 2.0Ω.