To solve the problem, we need to analyze the situation involving an electric cell, its internal resistance, and the external resistance connected to it. Let's break it down step-by-step. ### Given Information: - Nominal voltage of the cell, \( E \) - External resistance, \( R = 3 \, \Omega \) - Voltage across the external resistance when the current flows, \( V = 0.6E \) ### Step 1: Understanding the Circuit When a load (external resistance) is connected to an electric cell, the total voltage provided by the cell is divided between the internal resistance of the cell and the external resistance. The internal resistance of the cell is denoted as \( r \). ### Step 2: Applying Ohm's Law According to Ohm's Law, the voltage across a resistor is given by: \[ V = I \cdot R \] where \( I \) is the current flowing through the circuit. ### Step 3: Total Voltage in the Circuit The total voltage \( E \) of the cell is equal to the sum of the voltage across the external resistance and the voltage across the internal resistance: \[ E = V + V_{internal} \] where \( V_{internal} = I \cdot r \). ### Step 4: Expressing the Current From the information given, we know that when the voltage across the external resistance is \( 0.6E \), we can express the current \( I \) flowing through the circuit as: \[ V = I \cdot R \implies 0.6E = I \cdot 3 \] From this, we can solve for the current \( I \): \[ I = \frac{0.6E}{3} = 0.2E \] ### Step 5: Finding the Internal Resistance Now, we can substitute \( I \) back into the equation for the total voltage: \[ E = 0.6E + I \cdot r \] Substituting \( I = 0.2E \): \[ E = 0.6E + (0.2E) \cdot r \] Rearranging gives: \[ E - 0.6E = 0.2E \cdot r \] \[ 0.4E = 0.2E \cdot r \] Dividing both sides by \( 0.2E \) (assuming \( E \neq 0 \)): \[ \frac{0.4E}{0.2E} = r \implies 2 = r \] Thus, the internal resistance \( r \) of the cell is \( 2 \, \Omega \). ### Conclusion The correct answer is **B. 2Ω**. ### Explanation of Other Options: - **A. 1Ω**: This option is incorrect because it underestimates the internal resistance based on the voltage drop observed. - **C. 3Ω**: This option suggests that the internal resistance is equal to the external resistance, which would not allow for a voltage drop to \( 0.6E \) under the given conditions. - **D. 4Ω**: This option is incorrect as it overestimates the internal resistance, which would lead to a larger voltage drop than observed. ### Revision Summary: - The internal resistance of a cell can be calculated using the voltage drop across the external resistance and the nominal voltage. - Use Ohm's Law to relate voltage, current, and resistance in the circuit. - The total voltage is the sum of the voltage across the external resistance and the internal resistance. - Ensure to rearrange equations carefully to isolate the variable of interest (in this case, internal resistance).