A source delivers an AC voltage of the form Δv = (89.0 V) sin (70πt), where Δv is in volts and t is in seconds, to a capacitor. The maximum current in the circuit is 0.460 A. Find the following.
(a) Find the rms voltage of the source.
(b) Find the frequency of the source.
(c) Find the value of the capacitance.
To solve this problem, we will use the formulas and concepts related to AC circuits and capacitors.
(a) To find the rms voltage of the source, we need to determine the root mean square (rms) value of the given AC voltage. The rms value of an AC voltage is given by the formula:
Vrms = Vmax / √2
where Vmax is the maximum voltage amplitude.
In this case, the given voltage is Δv = (89.0 V) sin (70πt). Since the given value is the maximum voltage, we have Vmax = 89.0 V. Applying the formula, we get:
Vrms = 89.0 V / √2
Vrms ≈ 62.99 V
Therefore, the rms voltage of the source is approximately 62.99 V.
(b) The frequency of the source can be determined using the formula:
f = ω / (2π)
where ω is the angular frequency.
In this case, we are given that the voltage follows the equation Δv = (89.0 V) sin (70πt). By comparing this equation with the general form of a sinusoidal function, we can see that the angular frequency is 70π.
Using the formula, we have:
f = (70π) / (2π)
f = 35 Hz
Therefore, the frequency of the source is 35 Hz.
(c) To find the value of the capacitance, we can use the formula for capacitive reactance (Xc) in an AC circuit:
Xc = (1 / (2πfC))
where f is the frequency and C is the capacitance.
Given that the maximum current in the circuit is 0.460 A, we know that the current is equal to the maximum voltage divided by the impedance of the circuit (Z), which is the magnitude of the total impedance of the capacitor.
Since the current is given as the maximum value, we can convert it to the rms value using Irms = Imax / √2:
Irms = 0.460 A / √2
Irms ≈ 0.325 A
The impedance (Z) of a capacitor is given by the formula:
Z = Xc = Vrms / Irms
Substituting the values we have, we can solve for C:
Xc = (1 / (2πfC))
C = 1 / (2πfXc)
Plugging in the values, we get:
C = 1 / (2π * 35 * 0.325)
C ≈ 0.0018 F
Therefore, the value of the capacitance is approximately 0.0018 F.