An AC circuit has a voltage source 5.00cos(wt) V. There is also a 180. Ω resistor and a 2.70 nF capacitor in the circuit. What is the value of the peak voltages VR and VC if the emf frequency is 4.10 kHz?
Question
An AC circuit has a voltage source 5.00cos(wt) V. There is also a 180. Ω resistor and a 2.70 nF capacitor in the circuit. What is the value of the peak voltages VR and VC if the emf frequency is 4.10 kHz?
Solution
To find the peak voltages VR and VC, we first need to find the impedance (Z) of the circuit, the current (I) in the circuit, and the reactance (Xc) of the capacitor.
Step 1: Calculate the angular frequency (w) w = 2πf where f is the frequency. w = 2π * 4.10 kHz = 2π * 4100 Hz = 25736 rad/s
Step 2: Calculate the reactance (Xc) of the capacitor Xc = 1/(wC) where C is the capacitance. Xc = 1/(25736 rad/s * 2.70 nF) = 1/(25736 rad/s * 2.70*10^-9 F) = 14.4 Ω
Step 3: Calculate the impedance (Z) of the circuit Z = sqrt(R^2 + Xc^2) where R is the resistance. Z = sqrt((180 Ω)^2 + (14.4 Ω)^2) = 181 Ω
Step 4: Calculate the current (I) in the circuit I = V/Z where V is the peak voltage of the source. I = 5.00 V / 181 Ω = 0.0276 A
Step 5: Calculate the peak voltages VR and VC VR = IR VR = 0.0276 A * 180 Ω = 4.97 V
VC = IXc VC = 0.0276 A * 14.4 Ω = 0.40 V
So, the peak voltage across the resistor VR is 4.97 V and the peak voltage across the capacitor VC is 0.40 V.
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