The impedance of the circuit having a resistor of resistance 40Ω and a capacitor of capacitance 132.6F is 2.65Ω
The potential of the circuit = 120 V
The frequency of the circuit = 60 Hz
The capacitance of the capacitor = 132.6 F
The resistance of the resistor = 40Ω
The impedance of this circuit can be found using the formula
Z = R / (RωC + 1)
where Z is the impedance
R is the resistance of the resistor
C is the capacitance of the capacitor
ω is the angular frequency
First, let us find the angular frequency
ω = 1/2πf
where f is the frequency
ω = 1/ 2 x 3.14 x 60
= 1 / 376.8
= 0.00265
So, let us substitute the known values in the impedance formula, we get
Z = 40 / ( 40 x 0.00265 x 132.6 +1)
= 40 / (14.05 +1)
= 40 / 15.05
= 2.65Ω
Therefore, the impedance of the circuit is 2.65Ω
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