# Circuit help (parallel and series together)

1. Sep 2, 2014

### sloan13

1. The problem statement, all variables and given/known data
A 50V independent voltage source supplies power to three resistors in the circuit shown below. For each resistor find the voltage drop and current.

Circuit Diagram:
https://webwork2.uncc.edu/webwork2_files/tmp/Fall2014-Engr1201-Common/img/f65baf2e-12f5-3b7e-898f-18c0cd5a2145___71184e03-0c3d-3804-97e6-b93ed961be0a.gif [Broken]

RA = 55 Ω, RB = 310 Ω, and RC = 24 Ω.

Vdrop across RA = ____V
Vdrop across RB = ____V
Vdrop across RC = ____V

Current through RA = ____A
Current through RB = ____A
Current through RC = ____A

2. Relevant equations
V = RI

V = Voltage
R = Resistance
I = Current

____________________________

R = r1 + r2 + r3.... (series)
1/R = 1/r1 + 1/r2 + 1/r3... (parallel)

____________________________

3. The attempt at a solution
I have tried solving this by solving the voltage and current at each individual resistor and munipulating formulas. I haven't really gotten anywhere. We haven't been over this stuff in class and the handout doesn't cover this, but they refused to change the due date.

I am so lost.

Last edited by a moderator: May 6, 2017
2. Sep 2, 2014

### Staff: Mentor

Last edited by a moderator: May 6, 2017
3. Sep 2, 2014

### sloan13

Well, I have tried applying Kirchoff's Laws and tried to find the currents of the first two resistors, but I don't think I can without knowing all but one of them.

50V - IR1 - IR2 - IR3 = 0
50V - 55I1 - 310I2 - (24)(.707090275) = 0

I found the third current by taking the voltage/overall resistance. (The third current and voltage drop and they were correct)

I don't think I can calculate the voltage drop without knowing the current or vice versa.

4. Sep 2, 2014

### sloan13

Oh and I also tried getting the percentage of resistance of both sides of the parallel resistors and tried applying it to the current of both respectively.

5. Sep 2, 2014

### FOIWATER

Why not find the total current coming from the voltage supply? (simplify the resistor network) then you know the total current through Rc.

With the current through Rc, you can find the voltage drop across Rc. Then 50-Rc voltage drop is the voltage across both Ra and Rc. So you can find the current through each of those as well.

Does anything here not make sense?

6. Sep 2, 2014

### sloan13

Well I got Rc, but basically you are saying that I can find Ra voltage drop by:

50V = Ra - (Rc voltage drop)?

7. Sep 2, 2014

### FOIWATER

no Ra = 50- Rc voltage drop

50V would actually be the sum, that is: 50=(Ravoltage drop)+(Rcvoltage drop)

8. Sep 2, 2014

### FOIWATER

you know all the resistances so that's not an issue, the first trick is to find the total current from the source.

9. Sep 2, 2014

### FOIWATER

find an equivalent circuit resistance (the resistance seen by the source)

That is, Rc in series with (Ra and Rb in parallel)

then use ohms law to find total current. total current = 50v/total resistance

10. Sep 2, 2014

### sloan13

I got it! Thanks guys!

11. Sep 2, 2014

### sloan13

The issue for me was that I thought that 50V was the sum of the voltage of the entire circuit, as in Ra V + Rb V + Rc V. Again, thank you!

12. Sep 2, 2014

ahh, I see.

Most welcome