A 325-Ω resistor is connected in series with a 4.87-μF capacitor and a source of time-varying emf providing 79.5 V. (a) At what frequency will the potential drop across the capacitor equal that across the resistor? Hz (b) What is the current through the circuit when this occurs? A
Question
A 325-Ω resistor is connected in series with a 4.87-μF capacitor and a source of time-varying emf providing 79.5 V. (a) At what frequency will the potential drop across the capacitor equal that across the resistor? Hz (b) What is the current through the circuit when this occurs? A
Solution
(a) The potential drop across the capacitor equals that across the resistor when the impedance of the capacitor equals the resistance of the resistor. The impedance of a capacitor is given by the formula Zc = 1/(2πfC), where f is the frequency and C is the capacitance. Setting this equal to the resistance and solving for f gives:
325 Ω = 1/(2πf * 4.87 μF) => f = 1/(2π * 325 Ω * 4.87 μF) => f ≈ 1000 Hz
(b) The current through the circuit can be found using Ohm's law, I = V/R, where V is the voltage and R is the resistance. However, when the potential drop across the capacitor equals that across the resistor, the total voltage is divided equally between the two, so the voltage across the resistor (and thus the voltage used in Ohm's law) is half the total voltage:
I = (79.5 V / 2) / 325 Ω => I ≈ 0.122 A
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