# Complex Circuit Question

1. Aug 27, 2014

### pstir2

1. The problem statement, all variables and given/known data

Solve for all voltage drops and currents at the resistors and battery

2. Relevant equations
V=IR
RT=R1+R2+...+Rn
1/RT=1/R1+1/R2+...+1/Rn
IT=I1+I2+...+In

3. The attempt at a solution

I just wanted to check if my solution is correct

------Voltage----Current----Resistance
R1-----150--------3.75---------40
R2-----150--------3.75---------40
R3-----45---------0.75---------60
R4-----15-----------3------------5
R5-----30-----------2-----------15
R6-----20-----------1-----------20
R7-----10-----------1-----------10
Total---345--------3.75---------92

Last edited: Aug 27, 2014
2. Aug 27, 2014

### Staff: Mentor

This would be easier if the resistor numbers and current directions would be marked in the diagram...

You can simply check Kirchhoff's laws for every node and loop as cross-check.

3. Aug 27, 2014

### BvU

Is there something about V that you seem to know and we don't ?
So far I can only stamp an approval stamp on R6 and R7 (they happen to be the only ones with 10 and 20 Ohm).

4. Aug 27, 2014

### pstir2

Because the voltage drop though resistors R6(20ohms) and R7(10ohms) causes a total voltage drop of 30V any parallel circuits to it will also have a voltage drop of 30 V. therefore R5(15ohms) must have a voltage drop of 30V.

5. Aug 27, 2014

### pstir2

Because the voltage drop though resistors R6(20ohms) and R7(10ohms) causes a total voltage drop of 30V any parallel circuits to it will also have a voltage drop of 30 V. therefore R5(15ohms) must have a voltage drop of 30V.

6. Aug 27, 2014

### Staff: Mentor

The voltage drops of 30 V on the right side are fine, but the voltage of the voltage source is unclear.

7. Aug 27, 2014

### pstir2

The total is what I calculated the voltage source to be but I'm not sure its correct.

8. Aug 27, 2014

### Staff: Mentor

Which total?
You cannot solve this problem if you don't know the voltage drop at the voltage source.

9. Aug 27, 2014

### Staff: Mentor

It appears the ammeter is displaying a reading of 1.0A, meaning there is sufficient information to determine V.

10. Aug 28, 2014

### BvU

My mistake. I saw a current source instead of an ammeter.

11. Aug 28, 2014

### Staff: Mentor

Same here.
Okay, then it is fine.

12. Aug 28, 2014

### BvU

So herewith a stamp of approval goes to post #1. And now I run the risk of being chastized by the spirits that float over and guard PF: the forum isn't really intended for that kind of thing at all. With good reasons.
Still, it was nice to look into this exercise...