To solve the problem, we need to analyze the circuit based on the information provided about the current readings with different resistances. The circuit likely consists of a voltage source and a resistor \( R \) in series with an ammeter. ### Given: 1. When \( R = 5Ω \), the current \( I = 3A \). 2. When \( R = 2Ω \), the current \( I = 6A \). ### Step 1: Determine the Voltage in the Circuit Using Ohm's Law, which states that \( V = I \times R \), we can calculate the voltage across the circuit for both scenarios. **For \( R = 5Ω \) and \( I = 3A \):** \[ V = I \times R = 3A \times 5Ω = 15V \] **For \( R = 2Ω \) and \( I = 6A \):** \[ V = I \times R = 6A \times 2Ω = 12V \] ### Step 2: Analyze the Results We have two different voltage readings for the same circuit, which suggests that there is another resistor \( x \) in the circuit that affects the total resistance. ### Step 3: Set Up the Equations Assuming the circuit has a total voltage \( V \) that remains constant, we can express the total resistance in terms of \( x \) and the known resistances. 1. **For \( R = 5Ω \):** \[ V = I \times (R + x) \implies 15V = 3A \times (5Ω + x) \] Simplifying this gives: \[ 15 = 3(5 + x) \implies 15 = 15 + 3x \implies 3x = 0 \implies x = 0Ω \] (This is not a valid solution since \( x \) cannot be zero.) 2. **For \( R = 2Ω \):** \[ V = I \times (R + x) \implies 12V = 6A \times (2Ω + x) \] Simplifying this gives: \[ 12 = 6(2 + x) \implies 12 = 12 + 6x \implies 6x = 0 \implies x = 0Ω \] (Again, this is not valid.) ### Step 4: Re-evaluate the Circuit Since both calculations yield \( x = 0Ω \), we need to consider that the circuit might be more complex, possibly involving a parallel configuration or a different arrangement of resistors. ### Step 5: Solve for \( x \) Using the Two Scenarios We can set up a system of equations based on the two scenarios: 1. From the first scenario: \[ 15 = 3(5 + x) \implies 15 = 15 + 3x \implies 3x = 0 \implies x = 0Ω \] 2. From the second scenario: \[ 12 = 6(2 + x) \implies 12 = 12 + 6x \implies 6x = 0 \implies x = 0Ω \] ### Step 6: Find the Value of \( x \) To find \( x \), we can use the relationship between the two scenarios. Since the current changes with resistance, we can set up a ratio based on the currents and resistances: Using the formula for total resistance in series: \[ R_{total} = R + x \] From the two scenarios, we can derive: \[ \frac{I_1}{I_2} = \frac{R_2 + x}{R_1 + x} \] Substituting the values: \[ \frac{3}{6} = \frac{2 + x}{5 + x} \] Cross-multiplying gives: \[ 3(5 + x) = 6(2 + x) \] Expanding both sides: \[ 15 + 3x = 12 + 6x \] Rearranging gives: \[ 15 - 12 = 6x - 3x \implies 3 = 3x \implies x = 1Ω \] ### Conclusion The value of \( x \) is not among the options provided. However, if we consider the possibility of a miscalculation or misinterpretation of the circuit, we can conclude that the correct answer based on the calculations is \( x = 1Ω \). ### Revision Summary - Use Ohm's Law \( V = I \times R \) to find voltage across resistors. - Set up equations based on different scenarios to find unknowns. - Analyze the circuit configuration (series vs. parallel) to understand current flow. - Always check if the calculated values match the options provided in the question. In this case, the answer provided in the question (C. 10Ω) does not match our calculations, indicating a potential error in the question or options.