Question

A capacitor is connected across an ac generator whose frequency is755 Hz and whose peak output voltage is140 V. The rms current in the circuit is 3.0 A.

(a) What is the capacitance of thecapacitor?
----- F

Answer 1

The capacitance of the capacitor connected across an ac generator with a peak voltage of 140 V and a frequency of 755 Hz, which results in an rms current of 3.0 A, is approximately 6.37 × 10-6 F.

Step-by-step explanation:

To find the capacitance of the capacitor connected across the ac generator, we need to use the relationship between capacitive reactance (Xc), frequency (f), rms current (Irms), and rms voltage (Vrms). Since the generator provides a peak voltage of 140 V, we first need to convert this to rms voltage using the formula Vrms = Vpeak / √2, which gives us Vrms = 140 / √2 = 99 V approximately. Now, we can use the formula for capacitive reactance Xc = 1 / (2πfC) and the fact that Vrms = Irms Xc, to solve for the capacitance C.

The equation can be rearranged to solve for C as follows: C = 1 / (2πf Irms Xc). Since we have that Irms is 3.0 A, and Xc = Vrms / Irms = 99 V / 3.0 A = 33 Ω, we find the capacitance to be C = 1 / (2π(755 Hz)(33 Ω)), which calculates to approximately 6.37 × 10-6 F.