Please check my answers: Combined Circuit

  • #1
37
2

Homework Statement


1zqdpnd.jpg

ek492.jpg

10z8p02.jpg

Homework Equations




The Attempt at a Solution


Vt=30V
It=4.3A
Rt=7Ω

V1=12.9V
V2=2.15V
V3=2.15V
V4=4.3V
V5=12.9V
VA=4.3V
VB=4.3V

I1=4.3A
I2=2.15A
I3=2.15A
I4=2.15A
I5=4.3A
IA=2.15A
IB=4.3A

R1=3Ω
R2=1Ω
R3=1Ω
R4=2Ω
R5=3Ω
RA=2Ω
RB=1Ω

Did I break down the problem correctly? And are all these calculations correct?
 

Answers and Replies

  • #3
Charles Link
Homework Helper
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Homework Statement


1zqdpnd.jpg

ek492.jpg

10z8p02.jpg

Homework Equations




The Attempt at a Solution


Vt=30V
It=4.3A
Rt=7Ω

V1=12.9V
V2=2.15V
V3=2.15V
V4=4.3V
V5=12.9V
VA=4.3V
VB=4.3V

I1=4.3A
I2=2.15A
I3=2.15A
I4=2.15A
I5=4.3A
IA=2.15A
IB=4.3A

R1=3Ω
R2=1Ω
R3=1Ω
R4=2Ω
R5=3Ω
RA=2Ω
RB=1Ω

Did I break down the problem correctly? And are all these calculations correct?
It looks reasonably good, but electrical currents do have a direction=normally you mark the direction with an arrow and if the current goes opposite to the arrow, it is then a negative current. Also, the voltages you have are voltage drops across the resistors. Often the question will be what is the voltage at a point in the circuit, and in order to do this, a point in the circuit (e.g. the "-" terminal of the voltage source) will be designated as the "V=0" or "ground" point. Then the "+" terminal of the battery will be V=+30 V, and the voltages at other locations are determined from the voltage drops. All in all, you did reasonably good.
 

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