Review. The voltage phasor diagram for a certain series RLC circuit is shown in Figure P33.59. The resistance of the circuit is 75.0 Ω , and the frequency is 60.0 Hz. Find (a) the maximum voltage Δ V max , (b) the phase angle ϕ , (c) the maximum current, (d) the impedance, (e) the capacitance and (f) the inductance of the circuit, and (g) the average power delivered to the circuit.
Review. The voltage phasor diagram for a certain series RLC circuit is shown in Figure P33.59. The resistance of the circuit is 75.0 Ω , and the frequency is 60.0 Hz. Find (a) the maximum voltage Δ V max , (b) the phase angle ϕ , (c) the maximum current, (d) the impedance, (e) the capacitance and (f) the inductance of the circuit, and (g) the average power delivered to the circuit.
Review. The voltage phasor diagram for a certain series RLC circuit is shown in Figure P33.59. The resistance of the circuit is 75.0 Ω, and the frequency is 60.0 Hz. Find (a) the maximum voltage ΔVmax, (b) the phase angle ϕ, (c) the maximum current, (d) the impedance, (e) the capacitance and (f) the inductance of the circuit, and (g) the average power delivered to the circuit.
(a)
Expert Solution
To determine
The maximum voltage.
Answer to Problem 33.59AP
The maximum voltage is
22.4V.
Explanation of Solution
Given info: The resistance of the circuit is
75.0Ω, the frequency is
60.0Hz. The voltage drop across the resistor is
20.0V, the voltage drop across the inductor is
25.0V and the voltage drop across the capacitor is
15.0V.
The expression for maximum value of the voltage is,
ΔVmax=(ΔVR)2+(ΔVL−ΔVC)2
Here,
ΔVR is the voltage drop across the resistor.
ΔVL is the voltage drop across the inductor.
ΔVC is the voltage drop across the capacitor.
Substitute
20.0V for
ΔVR,
25.0V for
ΔVL and
15.0V for
ΔVC in the above expression.
ΔVmax=(20.0V)2+(25.0V−15.0V)2=22.36V≈22.4V
Conclusion:
Therefore, the maximum voltage is
22.4V.
(b)
Expert Solution
To determine
The phase angle.
Answer to Problem 33.59AP
The phase angle is
26.6°.
Explanation of Solution
Given info: The resistance of the circuit is
75.0Ω, the frequency is
60.0Hz. The voltage drop across the resistor is
20.0V, the voltage drop across the inductor is
25.0V and the voltage drop across the capacitor is
15.0V.
The expression for the phase angle is,
ϕ=sin−1(ΔVL−ΔVCΔVmax)
Substitute
22.4V for
ΔVmax,
25.0V for
ΔVL and
15.0V for
ΔVC in the above expression.
ϕ=sin−1(25.0V−15.0V22.4V)=26.5°≈26.6°
Conclusion:
Therefore, the phase angle is
26.6°.
(c)
Expert Solution
To determine
The maximum current.
Answer to Problem 33.59AP
The maximum current is
0.267A.
Explanation of Solution
Given info: The resistance of the circuit is
75.0Ω, the frequency is
60.0Hz. The voltage drop across the resistor is
20.0V, the voltage drop across the inductor is
25.0V and the voltage drop across the capacitor is
15.0V.
The expression for maximum current is,
Imax=ΔVRR
Here,
R is the resistance.
Substitute
20.0V for
ΔVR and
75.0Ω for
R in the above expression.
Imax=20.0V75.0Ω=0.267A
Conclusion:
Therefore, the maximum current is
0.267A.
(d)
Expert Solution
To determine
The impedance.
Answer to Problem 33.59AP
The impedance is
83.9Ω.
Explanation of Solution
Given info: The resistance of the circuit is
75.0Ω, the frequency is
60.0Hz. The voltage drop across the resistor is
20.0V, the voltage drop across the inductor is
25.0V and the voltage drop across the capacitor is
15.0V.
The expression for the impedance is,
Z=Rcosϕ
Substitute
75.0Ω for
R and
26.6° for
ϕ in the above expression.
Z=75.0Ωcos(26.6°)=83.87Ω≈83.9Ω
Conclusion:
Therefore, the impedance is
83.9Ω.
(e)
Expert Solution
To determine
The capacitance.
Answer to Problem 33.59AP
The capacitance is
47.2μF.
Explanation of Solution
Given info: The resistance of the circuit is
75.0Ω, the frequency is
60.0Hz. The voltage drop across the resistor is
20.0V, the voltage drop across the inductor is
25.0V and the voltage drop across the capacitor is
15.0V.
The circuit is series
RLC circuit, therefore the value of current throughout the circuit is same.
The expression capacitive reactance is,
XC=ΔVCImax
Substitute
15.0V for
ΔVC and
0.267A for
Imax in the above expression.
XC=15.0V0.267A=56.18Ω
The expression capacitive reactance in terms of the capacitance is,
XC=12πfC
Here,
f is the frequency.
C is the capacitance.
Rearrange the above equation for the value of capacitance.
C=12πfXC
Substitute
56.18Ω for
XC and
60.0Hz for
f in the above expression.
C=12π(60.0Hz)(56.18Ω)=47.2×10−6F=47.2μF
Conclusion:
Therefore, the capacitance is
47.2μF.
(f)
Expert Solution
To determine
The inductance of the circuit.
Answer to Problem 33.59AP
The inductance of the circuit
0.248H.
Explanation of Solution
Given info: The resistance of the circuit is
75.0Ω, the frequency is
60.0Hz. The voltage drop across the resistor is
20.0V, the voltage drop across the inductor is
25.0V and the voltage drop across the capacitor is
15.0V.
The circuit is series
RLC circuit, therefore the value of current throughout the circuit is same.
The expression inductive reactance is,
XL=ΔVLImax
Substitute
25.0V for
ΔVC and
0.267A for
Imax in the above expression.
XL=25.0V0.267A=93.63Ω
Thus the value of inductive reactance is
93.63Ω.
The expression inductive reactance in terms of the inductance is,
XL=2πfL
Here,
f is the frequency.
L is the capacitance.
Rearrange the above equation for the value of inductance.
L=XL2πf
Substitute
93.63Ω for
XC and
60.0Hz for
f in the above expression.
C=93.63Ω2π(60.0Hz)=0.248H≈0.248H
Conclusion:
Therefore, the inductance is
0.248H.
(g)
Expert Solution
To determine
The average power delivered to circuit.
Answer to Problem 33.59AP
The average power delivered to circuit is
2.67W.
Explanation of Solution
Given info: The resistance of the circuit is
75.0Ω, the frequency is
60.0Hz. The voltage drop across the resistor is
20.0V, the voltage drop across the inductor is
25.0V and the voltage drop across the capacitor is
15.0V.
The expression for R.M.S value of the current is,
IRMS=Imax2
The expression for the average power delivered is,
Pavg=(IRMS)2R
Substitute
Imax2 for
IRMS in the above equation.
Pavg=(Imax2)2R=(Imax)22R
Substitute
0.267A for
Imax and
75.0Ω for
R in the above equation.
Pavg=(0.267A)2275.0Ω=2.67W
Conclusion:
Therefore, the average power delivered to circuit is
2.67W.
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