To determine the internal resistance of the cell, we can use Ohm's Law and the concept of internal resistance in a circuit. Let's break down the problem step-by-step. ### Step 1: Understanding the Problem We have a cell that can supply different currents through two different resistors: - **Resistor 1 (R1)**: 4.0 Ω with a current of 0.4 A - **Resistor 2 (R2)**: 10.0 Ω with a current of 0.2 A The internal resistance of the cell (r) affects the total voltage supplied by the cell (V) when a load is connected. The total voltage can be expressed as: \[ V = I(R + r) \] where: - \( I \) is the current, - \( R \) is the external resistance, - \( r \) is the internal resistance of the cell. ### Step 2: Applying Ohm's Law For each resistor, we can express the voltage across the cell using the formula above. #### For Resistor 1 (4.0 Ω, 0.4 A): Using Ohm's Law: \[ V_1 = I_1(R_1 + r) \] Substituting the values: \[ V_1 = 0.4(4.0 + r) \] \[ V_1 = 1.6 + 0.4r \] (Equation 1) #### For Resistor 2 (10.0 Ω, 0.2 A): Similarly, we can write: \[ V_2 = I_2(R_2 + r) \] Substituting the values: \[ V_2 = 0.2(10.0 + r) \] \[ V_2 = 2.0 + 0.2r \] (Equation 2) ### Step 3: Equating the Voltages Since the voltage supplied by the cell is the same in both cases, we can set Equation 1 equal to Equation 2: \[ 1.6 + 0.4r = 2.0 + 0.2r \] ### Step 4: Solving for Internal Resistance (r) Now, we can solve for \( r \): 1. Rearranging the equation: \[ 0.4r - 0.2r = 2.0 - 1.6 \] \[ 0.2r = 0.4 \] 2. Dividing both sides by 0.2: \[ r = \frac{0.4}{0.2} = 2.0 \, \Omega \] ### Conclusion The internal resistance of the cell is **2.0 Ω**. Therefore, the correct option is **A. 2.0Ω**. ### Explanation of Other Options - **B. 1.0Ω**: This value would imply that the voltage drop across the internal resistance is much lower than what is observed, leading to incorrect calculations of the total voltage. - **C. 2.5Ω**: This value is higher than what we calculated, which would suggest an even larger voltage drop across the internal resistance, contradicting the observed currents. - **D. 1.5Ω**: Similar to option B, this value does not satisfy the equations derived from the observed currents and resistances. ### Revision Summary - Use Ohm's Law to relate current, voltage, and resistance in circuits. - The internal resistance can be calculated by equating the voltage expressions for different loads. - Ensure to isolate the internal resistance correctly when solving equations. - Always check the reasonableness of your answer against the context of the problem.