To solve the problem, we need to analyze the situation involving an electric cell with internal resistance and how it affects the voltage across an external resistor \( R \). Let's break it down step-by-step. ### Step 1: Understanding the Circuit We have an electric cell with: - Internal resistance \( r = 2 \, \Omega \) - The nominal voltage of the cell is not given, but we will denote it as \( V \). When a resistance \( R \) is connected across the terminals of the cell, the voltage across \( R \) is reduced to \( \frac{3}{5} \) of the nominal voltage \( V \). This means the voltage across \( R \) is: \[ V_R = \frac{3}{5} V \] ### Step 2: Applying Ohm's Law and the Voltage Divider Rule In a circuit with internal resistance, the total voltage \( V \) is divided between the internal resistance \( r \) and the external resistance \( R \). 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. The total voltage \( V \) can be expressed as: \[ V = I(R + r) \] where \( R \) is the external resistance and \( r \) is the internal resistance. ### Step 3: Setting Up the Equation From the information given, we know: \[ V_R = I \cdot R = \frac{3}{5} V \] Substituting \( V \) from the total voltage equation: \[ V_R = I \cdot R = \frac{3}{5} I(R + r) \] ### Step 4: Rearranging the Equation We can rearrange the equation to isolate \( R \): \[ I \cdot R = \frac{3}{5} I(R + r) \] Assuming \( I \neq 0 \) (since there is current flowing), we can divide both sides by \( I \): \[ R = \frac{3}{5}(R + r) \] ### Step 5: Solving for \( R \) Now, we can distribute \( \frac{3}{5} \): \[ R = \frac{3}{5}R + \frac{3}{5}r \] To isolate \( R \), we can subtract \( \frac{3}{5}R \) from both sides: \[ R - \frac{3}{5}R = \frac{3}{5}r \] This simplifies to: \[ \frac{2}{5}R = \frac{3}{5}r \] Now, multiply both sides by \( \frac{5}{2} \) to solve for \( R \): \[ R = \frac{3}{2}r \] ### Step 6: Substituting the Value of \( r \) Now, substitute \( r = 2 \, \Omega \): \[ R = \frac{3}{2} \times 2 = 3 \, \Omega \] ### Conclusion Thus, the value of \( R \) is \( 3 \, \Omega \). Therefore, the correct option is: **A. 3 Ω** ### Explanation of Other Options - **B. 2 Ω**: This would imply that the external resistance is equal to the internal resistance, which would not lead to a reduction of voltage to \( \frac{3}{5} \) of the nominal value. - **C. 1 Ω**: This value is too low and would not satisfy the condition of the voltage drop being \( \frac{3}{5} \) of the nominal voltage. - **D. 6 Ω**: This value is too high and would lead to a larger voltage drop than what is specified in the problem. ### Revision Summary - The internal resistance affects the total voltage across the external resistor. - The voltage across the external resistor can be expressed in terms of the total voltage and the internal resistance. - The relationship between the external resistance and internal resistance can be derived using Ohm's Law. - The final value of the external resistance \( R \) is \( 3 \, \Omega \).