Answer was given. Solve properly.Id = 2.4157 mAVgs = -1.8017 VVd = 10.2023 VVs = 3.6236 VVds = 6.5787 VVdg = 8.3795 V Determine the following for thenetwork shown:a. lpb. VGSc. Vpd. Vse. Vpsf. VDGV₁oC₁HF5 μFR 2 2.1 ΜΩR2 – 270 ΚΩGRDRs• 2.4 ΚΩDSH1₁O +16 V10 μFHe₂¹Dss = 8 mAVp=-4 V1.5 k C 20 μF

We need to find out current and voltage for given circuit

Determine the following for the network shown: a. l b. VGS c. Vp d. Vs e. Vps f. VDG V₁0- C₁ HF 5 μF R, 22.1 ΜΩ w R₂ 270 km2 Rp2.4 k D S +16 V + 10 μF Hc₂ ¹Dss=8mA Vp=-4 V Rs1.5 k C 20 μF

Determine the following for the network shown: a. l b. VGS c. Vp d. Vs e. Vps f. VDG V₁0- C₁ HF 5 μF R, 22.1 ΜΩ w R₂ 270 km2 Rp2.4 k D S +16 V + 10 μF Hc₂ ¹Dss=8mA Vp=-4 V Rs1.5 k C 20 μF

EBK ELECTRICAL WIRING RESIDENTIAL

19th Edition

ISBN:9781337516549

Author:Simmons

Publisher:Simmons

Chapter19: Special-purpose Outlets-water Pump, Water Heater

Section19.2: Water Heater Branch Circuit

Problem 8R: For residential water heaters, the Consumer Product Safety Commission suggests a maximum temperature...

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Solenoidal Vector Field

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Question

Answer was given. Solve properly.

Id = 2.4157 mA

Vgs = -1.8017 V

Vd = 10.2023 V

Vs = 3.6236 V

Vds = 6.5787 V

Vdg = 8.3795 V

Determine the following for the
network shown:
a. lp
b. VGS
c. Vp
d. Vs
e. Vps
f. VDG
V₁o
C₁
HF
5 μF
R 2 2.1 ΜΩ
R2 – 270 ΚΩ
G
RD
Rs
• 2.4 ΚΩ
D
S
H1₁
O +16 V
10 μF
He₂
¹Dss = 8 mA
Vp=-4 V
1.5 k C 20 μF

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