Final answer:
The percent increase in current when a 1.00-MΩ voltmeter is in parallel with a 75.0-kΩ resistor is approximately 1.33%, with the correct answer being B) 1.33%.
Step-by-step explanation:
The student's question relates to a circuit with a voltmeter in parallel with a resistor and how that affects the current flow. First, we need to find the equivalent resistance of a 1.00-MΩ voltmeter in parallel with a 75.0-kΩ resistor. The formula for equivalent resistance (R_eq) in parallel is 1/R_eq = 1/R1 + 1/R2. For our values:
1/R_eq = 1/1,000,000 Ω + 1/75,000 Ω ≈ 1/1,000,000 Ω + 0.0000133 Ω⁻¹ = 0.0000143 Ω⁻¹
So, R_eq ≈ 1/0.0000143 Ω⁻¹ ≈ 69930 Ω or 69.93 kΩ.
Now, to find the percent increase in current, we assume the voltage across them remains the same. Before, the current (I_initial) was V/R_initial, and now it is V/R_eq. The increase in current (ΔI) is then (V/R_eq) - (V/R_initial). The percent increase in current = (ΔI/I_initial) * 100%. Substituting the resistances, we get:
Percent increase = [(V/69.93 kΩ) - (V/75 kΩ)] / (V/75 kΩ) * 100% ≈ 1.33%.
Therefore, the correct answer to the student's question is B) 1.33%.