Please check my answers: Combined Circuit

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The calculations for the combined circuit homework appear to be mostly correct, with the total voltage (Vt=30V), total current (It=4.3A), and total resistance (Rt=7Ω) accurately stated. The voltage drops across the resistors and the currents through each branch are also correctly identified. However, it's important to note that electrical currents have a direction, which should be marked with arrows, and any current flowing opposite to this direction should be considered negative. Additionally, when determining voltages at specific points in the circuit, a reference point, often designated as ground (V=0), should be established to calculate other voltages accurately. Overall, the breakdown of the problem is sound, with minor adjustments needed for clarity on current direction and voltage reference points.
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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?
 
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Yes, it seems correct.
 
j doe said:

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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