I.SOLVE THE TOTAL RESISTANCE, R7. (SERIES-PARALLEL)II.SOLVE THE TOTAL CURRENT, I7. (OHM'S LAW)SHOW THAT VR1 + VR4 = 110 V. (VOLTAGE DIVIDER)SOLVE THE CURRENT IN R4, I. (CURRENT DIVIDER)V.III.%3DIV.SHOW ALL COMPUTATIONS.R, = 60 N%3DR2 = 40 2 R3 = 20 2wwww110 VR4 = 50 NITI (in R4)Vs(V)110VR1VR4VRI+VR4(OHM)(A)COMPUTEDMEASURED110|% ERROR

As per our company guidelines we are supposed to answer only first 3 subpart kindly repost the other…

I. SOLVE THE TOTAL RESISTANCE, RȚ. (SERIES-PARALLEL) II. SOLVE THE TOTAL CURRENT, I7. (OHM'S LAW) III. SHOW THAT Vr1+ VR4 = 110 V. (VOLTAGE DIVIDER) %3D SOL U T TIE CT TRD ENT D ND T TD ENT DVIT DED)

I. SOLVE THE TOTAL RESISTANCE, RȚ. (SERIES-PARALLEL) II. SOLVE THE TOTAL CURRENT, I7. (OHM'S LAW) III. SHOW THAT Vr1+ VR4 = 110 V. (VOLTAGE DIVIDER) %3D SOL U T TIE CT TRD ENT D ND T TD ENT DVIT DED)

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter7: Parallel Circuits

Section: Chapter Questions

Problem 3PP: Using the rules for parallel circuits and Ohmslaw, solve for the missing values....

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Concept explainers

Power Measurement

According to Blondel's theorem, the total power is measured with one less number of wattmeters than the number of wires present in the system. So, a single-phase and two-wire system will require one wattmeter and a single-phase and the three-wire system will require two wattmeters, and a three-phase and the three-wire system will also require two wattmeters and a three-phase and the four-wire system will require three wattmeter devices. The wattmeter is a device that measures power using one voltage and one current input as shown in the above figure. Multiple current or voltage input pair cables are used by many power analyzers and digital storage oscilloscopes (DSOs) for measuring watts, in effect within a single instrument acting as multiple wattmeters. It is possible to measure three-phase and four-wire power with one correctly specified power analyzer.

Question

100%

I.
SOLVE THE TOTAL RESISTANCE, R7. (SERIES-PARALLEL)
II.
SOLVE THE TOTAL CURRENT, I7. (OHM'S LAW)
SHOW THAT VR1 + VR4 = 110 V. (VOLTAGE DIVIDER)
SOLVE THE CURRENT IN R4, I. (CURRENT DIVIDER)
V.
III.
%3D
IV.
SHOW ALL COMPUTATIONS.
R, = 60 N
%3D
R2 = 40 2 R3 = 20 2
wwww
110 V
R4 = 50 N
IT
I (in R4)
Vs
(V)
110
VR1
VR4
VRI+VR4
(OHM)
(A)
COMPUTED
MEASURED
110
|% ERROR

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