Engineering Circuit help (parallel and series together)

[sloan13]

Homework Statement

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

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

Homework Equations

V = RI

V = Voltage
R = Resistance
I = Current

____________________________

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

____________________________

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.

 

[berkeman]

Homework Statement

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

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

Homework Equations

V = RI

V = Voltage
R = Resistance
I = Current

____________________________

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

____________________________

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.

Please show us your work so far. It's pretty hard to help you without seeing how you are approaching the problem.

 

[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.

 

[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.

 

[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?

 

[sloan13]

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

50V = Ra - (Rc voltage drop)?

 

[FOIWATER]

no Ra = 50- Rc voltage drop

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

 

[FOIWATER]

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

 

[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

 

[sloan13]

I got it! Thanks guys!

 

[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!

 

[FOIWATER]

ahh, I see.

Most welcome