PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 21, Problem 37P

(a)

To determine

The voltages VL,VC,VR, and the phase angle.

(a)

Expert Solution
Check Mark

Answer to Problem 37P

The values of the voltages are VL=3.4V_, VC=9.2V_, VR=6.9V_, and the phase angle is ϕ=40°_.

Explanation of Solution

Given that, in the series RLC circuit, the capacitance is 0.20mF, the inductance is 13mH, the resistance is 10.0Ω, the voltage amplitude is 9.0V, and the frequency is 60Hz. The circuit is shown in Figure 1.

PHYSICS, Chapter 21, Problem 37P , additional homework tip  1

Write the expression for the phase angle in the series RLC circuit.

ϕ=tan1XLXCR (I)

Here, ϕ is the phase angle, XL is the inductive reactance, XC is the capacitive reactance, and R is the resistance.

Write the expression for the inductive reactance.

XL=2πfL (II)

Here, f is the frequency, and L is the inductance.

Write the expression for the capacitive reactance.

XC=12πfC (III)

Here, C is the capacitance.

Write the expression for the impedance of the series RLC circuit.

Z=Rcosϕ (IV)

Here, Z is the impedance of the circuit.

Write the expression for the current in the series RLC circuit.

I=εmZ (V)

Here, I is the current, and εm is the voltage amplitude.

Write the expression for the voltage across the inductor (VL).

VL=IXL (VI)

Write the expression for the voltage across the capacitor (VC).

VC=IXC (VII)

Write the expression for the voltage across the resistor (VR).

VR=IR (VIII)

Conclusion:

Substitute 60Hz for f, and 13mH for L in equation (II) to find XL.

XL=2π(60Hz)(13mH)=2π(60Hz)(13mH×1H1000mH)=4.901Ω

Substitute 60Hz for f, and 0.20mF for C in equation (III) to find XC.

XC=12π(60Hz)(0.20mF)=12π(60Hz)(0.20mF×1F1000mF)=13.26Ω

Substitute 4.901Ω for XL, 13.26Ω for XC, and 10.0Ω for R in equation (I) to find ϕ.

ϕ=tan14.901Ω13.26Ω10.0Ω=39.9°40°

Substitute 39.9° for ϕ, and 10.0Ω for R in equation (IV) to find Z.

Z=10.0Ωcos(39.9°)=13.035Ω

Substitute 13.035Ω for Z, and 9.0V for εm in equation (V) to find I.

I=9.0V13.035Ω=0.6904A

Substitute 0.6904A for I, and 4.901Ω for XL in equation (VI) to find VL.

VL=(0.6904A)(4.901Ω)=3.4V

Substitute 0.6904A for I, and 13.26Ω for XC in equation (VII) to find VC.

VL=(0.6904A)(13.26Ω)=9.2V

Substitute 0.6904A for I, and 10.0Ω for R in equation (VIII) to find VR.

VL=(0.6904A)(10.0Ω)=6.9V

Therefore, the values of the voltages are VL=3.4V_, VC=9.2V_, VR=6.9V_, and the phase angle is ϕ=40°_.

(b)

To determine

The phasor diagram of the voltages of the given RLC circuit.

(b)

Expert Solution
Check Mark

Answer to Problem 37P

The phasor diagram of the voltages of the given RLC circuit is shown in Figure 2.

Explanation of Solution

For the given series RLC circuit I Figure 1, the values of the voltages obtained as VL=3.4V, VC=9.2V, VR=6.9V, and the phase angle is ϕ=40°.

The difference of the voltages across the inductor and capacitor is obtained to be,

VLVC=3.4V9.2V=5.8V

Conclusion:

The phase difference between the voltages VL and VC is πrad, and the phase difference between VL and VR is π/2rad. The phasor diagram of the voltages of the given RLC circuit is shown in Figure 2.

PHYSICS, Chapter 21, Problem 37P , additional homework tip  2

Therefore, the phasor diagram of the voltages of the given RLC circuit is shown in Figure 2.

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Chapter 21 Solutions

PHYSICS

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