The three-phase source line-to-neutral voltages are given by E a n = 10 ∠ 0 ° , E b h = 10 ∠ + 240 ° , and E c n = 10 ∠ − 240 ° volts . Is the source balanced? (a) Yes (b) No
The three-phase source line-to-neutral voltages are given by E a n = 10 ∠ 0 ° , E b h = 10 ∠ + 240 ° , and E c n = 10 ∠ − 240 ° volts . Is the source balanced? (a) Yes (b) No
Solution Summary: The author explains that the sum of voltage across all three phases is equal to zero for the three-phase balanced source.
Find the operating point and the load line of a voltage-divider JFET biasing circuit
using the following parameters: VGS(0) = -1.3 and Vcc = 15 volts. Assume
ipss = 20 mA, RG₁ = RG2 = 10 kn, RD = 300, and Rs = 1 kn. Use Fig. 4b for
the IV characteristic of the JFET.
20nA
GS=-1.3 GS
10nA-
50
100
150
200
ID(J1)
UDS
Fig. 4b. The IV characteristics of an n-channel JFET (J113). The plots are for VGs increments of
0.05 volts. VGS(0) -1.3. The yellow and blue load lines are for examples 2 &3,
respectively.
Design the JFET circuit for the largest in swing. Use the self-bias circuit shown in
Fig. 6. Assume that VGS (0) = -1.3 and Vcc = 15 volts. Furthermore, assume that
ipss = 20 mA. Using Fig. 4b, draw the load line and identify the Q point. Explain
why this will allow the largest swing. Use ip = ipss (1-
VGS
VGS(0)
to show what
happens to i, and vps when you have a swing of 0.2 volts in vcs form its operating
point (that is, change vas by ±0.2 volts and compute the corresponding
iD and VDs).
RD
RG
Rs
0
20nA
GS=-1.3 VGS
12
10nA
-0-
Fig. 6. Circuit for Examples 2 &3.
BA-C
50
100
150
200
□ ID(J1)
UDS
Fig. 4b. The IV characteristics of an n-channel JFET (J113). The plots are for VGs increments of
0.05 volts. VGS(0) -1.3. The yellow and blue load lines are for examples 2 &3,
respectively.
please do the correct VI chrastaristics curve on excel. I am not sure if mine is correct
Chapter 2 Solutions
MindTap Engineering for Glover/Overbye/Sarma's Power System Analysis and Design, 6th Edition, [Instant Access], 1 term (6 months)
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