Design and simulate the circuit in figure 2 using Multisim software and get the following parameters  Current in circuit I =…………………………mA  Voltage across R1 ER =…………………………V  Voltage across L1 EL =…………………………V  Power dissipated P =…………………………mW

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.7P: Let a 100V sinusoidal source be connected to a series combination of a 3 resistor, an 8 inductor,...
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Design and simulate the circuit in figure 2 using Multisim software and get the following parameters

 Current in circuit I =…………………………mA
 Voltage across R1 ER =…………………………V
 Voltage across L1 EL =…………………………V
 Power dissipated P =…………………………mW

When an ac voltage is applied across a pure inductance, the current lags the voltage by
90°. Inductance therefore has a phase angle associated with it. The opposition that an
inductance offers to the flow of alternating current is called inductive reactance XL and
the impedance of inductor can be expressed as XL<90°, or jXL. The magnitude of XL is
XL=2TfL= wL. It implies the impedance of inductor as,
ZL = jwL = wL < (90°)
An RL series circuit with an ac supply voltage is shown in below circuit. The impedance
of this circuit can be expressed as
ZT = ZR + ZL = R<0° + XL<+90°
The current in the circuit is
1 = E / ZT (The current lags the voltage)
The voltage across R is
VR = I. ZR
The voltage across L is
VL = I. ZL
Transcribed Image Text:When an ac voltage is applied across a pure inductance, the current lags the voltage by 90°. Inductance therefore has a phase angle associated with it. The opposition that an inductance offers to the flow of alternating current is called inductive reactance XL and the impedance of inductor can be expressed as XL<90°, or jXL. The magnitude of XL is XL=2TfL= wL. It implies the impedance of inductor as, ZL = jwL = wL < (90°) An RL series circuit with an ac supply voltage is shown in below circuit. The impedance of this circuit can be expressed as ZT = ZR + ZL = R<0° + XL<+90° The current in the circuit is 1 = E / ZT (The current lags the voltage) The voltage across R is VR = I. ZR The voltage across L is VL = I. ZL
By Kirchhoffs voltage law, then
[V=E-VR-VL = 0
E = VR + VL
EA
9Vrms
50Hz
0°
ww
2kQ
Figure 2: RL Circuit
L1
500mH
Transcribed Image Text:By Kirchhoffs voltage law, then [V=E-VR-VL = 0 E = VR + VL EA 9Vrms 50Hz 0° ww 2kQ Figure 2: RL Circuit L1 500mH
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Follow-up Question

Draw power triangle and  phasor diagram of the circuit

Compare the values measured in software analysis and calculated in mathematical
analysis and discuss the result obtained

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Follow-up Question

Calculate the following values using circuit analysis with detailed steps of calculation.


Reactance of L1 XL = …………..Ω
Total Impedance ZT = …………..Ω
Current in circuit I =……………mA
Figure 2: RL Circuit
Voltage across R1 ER =……………V
Voltage across L1 EL =……………V
Power dissipated P =……………W
Quality Factor Q.F = XL /R =………
Phase angle θ =……………
Total reactive power in the circuit Q =……………VAr
Apparent power delivered by source S =…………….VA

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Follow-up Question

Observe the following waveforms using oscilloscope
 Source voltage and source current waveforms
 Voltage across resistor
 Voltage across inductor

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