Design and Simulate the circuit in figure 1 using Multisim software and get the following parameters  Current in circuit I =…………………………mA  Voltage across R8 ER =…………………………V  Voltage across C2 EC =…………………………V  Power dissipated P =…………………………mW

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Design and Simulate the circuit in figure 1 using Multisim software and get the following parameters
 Current in circuit I =…………………………mA
 Voltage across R8 ER =…………………………V
 Voltage across C2 EC =…………………………V
 Power dissipated P =…………………………mW 

Objective
1. To understand the characteristics of a RC series network in ac circuit.
2. To understand the capacitive reactance, impedance, and phase angle.
Theory of operation
When an ac voltage is applied across a pure resistance, the resultant current is in
phase with the applied voltage. Resistance therefore has no phase angle associated
with it and impedance of resistor is simply written as R<0°. When an ac voltage is
applied across a pure capacitor, the resultant current leads the voltage by 90°.
Capacitance therefore has a phase angle associated with it. The opposition that a
capacitor offers to the flow of alternating current is called capacitive reactance Xc and
the impedance of a capacitor is written as Zc = Xc< -90°, or -jXc. The magnitude of Xc is
Xc = 1/(2TfC) = 1/wC. It implies, the impedance of capacitor is,
An RC series circuit with an ac supply voltage is shown below circuit. The impedance of
this circuit can be expressed as
ZT = ZR + Zc = R<0° + Xc<-90°
The current in the circuit is
Zc = = < (-90°)
jwC
WC
I= E/ZT
The voltage across R is
VR = I. ZR
The voltage across C is
Vc= I. Zc
By Kirchchoff's voltage law, then
EV = E-VR-Vc=0
Transcribed Image Text:Objective 1. To understand the characteristics of a RC series network in ac circuit. 2. To understand the capacitive reactance, impedance, and phase angle. Theory of operation When an ac voltage is applied across a pure resistance, the resultant current is in phase with the applied voltage. Resistance therefore has no phase angle associated with it and impedance of resistor is simply written as R<0°. When an ac voltage is applied across a pure capacitor, the resultant current leads the voltage by 90°. Capacitance therefore has a phase angle associated with it. The opposition that a capacitor offers to the flow of alternating current is called capacitive reactance Xc and the impedance of a capacitor is written as Zc = Xc< -90°, or -jXc. The magnitude of Xc is Xc = 1/(2TfC) = 1/wC. It implies, the impedance of capacitor is, An RC series circuit with an ac supply voltage is shown below circuit. The impedance of this circuit can be expressed as ZT = ZR + Zc = R<0° + Xc<-90° The current in the circuit is Zc = = < (-90°) jwC WC I= E/ZT The voltage across R is VR = I. ZR The voltage across C is Vc= I. Zc By Kirchchoff's voltage law, then EV = E-VR-Vc=0
EA
9Vrms
50Hz
0°
R1
2kQ
Figure 1: RC Circuit
C1
=47μF
Transcribed Image Text:EA 9Vrms 50Hz 0° R1 2kQ Figure 1: RC Circuit C1 =47μF
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Follow-up Question

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

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Follow-up Question

Draw power triangle of circuit
Draw the phasor diagram of the circuit

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Follow-up Question

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


Reactance of C2 Xc = …………..Ω
Total Impedance ZT = …………..Ω
Current in circuit I =……………mA
Voltage across R8 ER =……………V
Voltage across C2 EC =……………V
Power dissipated P =……………W
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 capacitor

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