Nt1310 Week 4 Lab Report
Assignment Week 4
Chapter 11
20. A parallel RL circuit has values of IL = 240 mA and IR = 180 mA. Calculate the magnitude and phase angle of the circuit current.
It=√ IL^2 + IR^2= √ 240mA^2 + 180mA^2= 0.3 x 10^-3= 300mA θ= tan^-1(-IL/IR)= (-240mA/180mA)= -53°
300mA∟-53°
22. Calculate the total impedance for the circuit shown in Figure 11.38b.
It= √IL^2 + IR^2= √800mA^2 + 120mA^2= 0.808949 x 10^-3= 808.95mA
Zt= Vs/It= 120V/808.95mA = 148Ω θ= tan^-1(-IL/IR)= (-800mA/120mA)= -81.5= 82°
148∟82°
24. Calculate the magnitude and phase angle of the circuit impedance shown in Figure 11.39b.
IL= Vs/XL= 4V/38Ω= 0.10526 x 10^-3= 105.26mA
IR= Vs/R= 4V/240Ω= 0.01666 x 10^-3= 16.67mA
IT= √ IL^2 + IR^2= √ 105.26mA^2 + 16.67mA^2= 0.10657 x 10^-3= 106.57mA θ= tan^-1(-IL/IR)= …show more content…
The voltage source in Figure 11.42b has the frequency limits shown. Calculate the circuit current and impedance values at the frequency limits of the source.
When f= 4kHz
XL= 2πfL= 2π x 4kHz x 47mH= 1181 x 10^3= 1.18kΩ
IL= Vs/XL= 20V/1.18k= 0.016949 x 10^-3= 16.95mA
IR= Vs/R= 20V/3.3k= 0.006060 x 10^-3= 6.06mA
IT= √ IL^2 + IR^2= √16.95mA^2 + 6.06mA^2= 0.01800 x 10^-3= 18mA θ= tan^-1(-IL/IR)= (-18mA/6.06mA)= -71°
Zt= Vs/It= 20V/18mA= 1111 x 10^3= 1.1kΩ
When f= 11kHx
XL= 2πfL= 2π x 11kHz x 47mH= 3248 x 10^3= 3.25kΩ
IL= Vs/XL= 20V/3.25k= 0.0061538 x 10^-3= 6.15mA
IR= Vs/R= 20V/3.3k= 0.006060 x 10^-3= 6.06mA
IT= √ IL^2 + IR^2= √6.15mA^2 + 6.06mA^2= 0.008634 x 10^-3= 8.63mA θ= tan^-1(-IL/IR)= (-6.15mA/6.06mA)= -45.57= -45.6°
Zt= Vs/It= 20V/8.63mA= 2317 x 10^3= 2.32kΩ
32. Refer to Problem 24. Express the circuit impedance in rectangular form.
37.53Ω∟81°
37.53 (cos 81°)= 37.53(0.156)= 5
37.53 (sin 81°)= 37.53(0.988)= 37.08Ω
5 + j37.08Ω
Chapter 13
22. A parallel RC circuit has values of IC = 320 mA and IR = 140 mA. Calculate the magnitude and phase angle of the circuit current.
It=√ IC^2 + IR^2= √ 320mA^2 + 140mA^2= 0.34928 x 10^-3= 349.28mA θ= tan^-1(IC/IR)= (320mA/140mA)= 66.41°
Chapter 11
20. A parallel RL circuit has values of IL = 240 mA and IR = 180 mA. Calculate the magnitude and phase angle of the circuit current.
It=√ IL^2 + IR^2= √ 240mA^2 + 180mA^2= 0.3 x 10^-3= 300mA θ= tan^-1(-IL/IR)= (-240mA/180mA)= -53°
300mA∟-53°
22. Calculate the total impedance for the circuit shown in Figure 11.38b.
It= √IL^2 + IR^2= √800mA^2 + 120mA^2= 0.808949 x 10^-3= 808.95mA
Zt= Vs/It= 120V/808.95mA = 148Ω θ= tan^-1(-IL/IR)= (-800mA/120mA)= -81.5= 82°
148∟82°
24. Calculate the magnitude and phase angle of the circuit impedance shown in Figure 11.39b.
IL= Vs/XL= 4V/38Ω= 0.10526 x 10^-3= 105.26mA
IR= Vs/R= 4V/240Ω= 0.01666 x 10^-3= 16.67mA
IT= √ IL^2 + IR^2= √ 105.26mA^2 + 16.67mA^2= 0.10657 x 10^-3= 106.57mA θ= tan^-1(-IL/IR)= …show more content…
The voltage source in Figure 11.42b has the frequency limits shown. Calculate the circuit current and impedance values at the frequency limits of the source.
When f= 4kHz
XL= 2πfL= 2π x 4kHz x 47mH= 1181 x 10^3= 1.18kΩ
IL= Vs/XL= 20V/1.18k= 0.016949 x 10^-3= 16.95mA
IR= Vs/R= 20V/3.3k= 0.006060 x 10^-3= 6.06mA
IT= √ IL^2 + IR^2= √16.95mA^2 + 6.06mA^2= 0.01800 x 10^-3= 18mA θ= tan^-1(-IL/IR)= (-18mA/6.06mA)= -71°
Zt= Vs/It= 20V/18mA= 1111 x 10^3= 1.1kΩ
When f= 11kHx
XL= 2πfL= 2π x 11kHz x 47mH= 3248 x 10^3= 3.25kΩ
IL= Vs/XL= 20V/3.25k= 0.0061538 x 10^-3= 6.15mA
IR= Vs/R= 20V/3.3k= 0.006060 x 10^-3= 6.06mA
IT= √ IL^2 + IR^2= √6.15mA^2 + 6.06mA^2= 0.008634 x 10^-3= 8.63mA θ= tan^-1(-IL/IR)= (-6.15mA/6.06mA)= -45.57= -45.6°
Zt= Vs/It= 20V/8.63mA= 2317 x 10^3= 2.32kΩ
32. Refer to Problem 24. Express the circuit impedance in rectangular form.
37.53Ω∟81°
37.53 (cos 81°)= 37.53(0.156)= 5
37.53 (sin 81°)= 37.53(0.988)= 37.08Ω
5 + j37.08Ω
Chapter 13
22. A parallel RC circuit has values of IC = 320 mA and IR = 140 mA. Calculate the magnitude and phase angle of the circuit current.
It=√ IC^2 + IR^2= √ 320mA^2 + 140mA^2= 0.34928 x 10^-3= 349.28mA θ= tan^-1(IC/IR)= (320mA/140mA)= 66.41°