simple resistor problem (some in series and in parallel)

Thread Starter

Hawkeye87

Joined Oct 7, 2008
52
The resistor that is blank is 6.1 ohms. The voltage is 12V. The questions are to find the current through each resistor. The 4.5 resistor i found the current came out to be 2.667A but it says it's wrong. What is it i'm doing wrong? I found the Rt to be 1.729 Ω. Is that correct?
 

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AlexK

Joined May 23, 2007
34
I think you have miscalculated Rt, can you tell us how you found it?

And even if Rt=1.729 Ω is correct, the calculation of the current through the 4.5 Ω resistor is incorrect. Can you show how you did that as well?
 

thatoneguy

Joined Feb 19, 2009
6,359
Loop analysis, but here's the "talk version"

(yes, electrons go from negative to postive, but if you stick with one convention, the math works out the same)

Current goes through the first two resistors in series, (5.5 Ω), then the 3.2 Ω in parallel with the sum of R (6.1) and 5.8 Ω, returning back to the source.

Hint: The current going through the 4.5 ohm and 1 ohm resistor will be the same, since there isn't another path. Likewise, the current through the 5.8 ohm and the R(6.1 ohm) will be the same.

I'll stop there, since it looks like a quiz. :)
 

pkennedy

Joined Feb 27, 2009
44
one way to think abouy it current remains constant in a series circuit and voltage remains constant in a parallel circuit maybe check out ohms law and kirchoffs law
 

Thread Starter

Hawkeye87

Joined Oct 7, 2008
52
I think you have miscalculated Rt, can you tell us how you found it?

And even if Rt=1.729 Ω is correct, the calculation of the current through the 4.5 Ω resistor is incorrect. Can you show how you did that as well?
Ok.....i found Rt like this.....
I added the 4.5 and the 1.0 resistor in series......I added the 6.1 resistor and the 5.8 resistor in series.......Then i added the Req1 and the Req2 all in parallel with the 3.2 resistor. Then to find the current through the 4.5 resistor i used the current divider, after finding I(t). I've taken an advanced circuits class but that was last semester and i gave my friend all my work and notes and i'm only working off of my knowledge and my friends book, which happens to be a physics book......go figure.
 

thatoneguy

Joined Feb 19, 2009
6,359
Rounded off, I have source current of ≈1.5 A and RTH of ≈ 8 Ω


\(R_{TH}=4.5+1.0+(\frac{1}{\frac{1}{3.2} +\frac{1}{6.1 +5.8}})\)

Divide Voltage by Resistance for current. To 4 digits, I arrived at, I = 1.4959 A, RTH = 8.0219 Ω

Multiply 1.4959 A * 5.5 Ω for voltage drop across first resistors, subtract that from 12V, then calculate current through the two parallel resistors. Using voltage from previous step. The currents in the parallel resistors need to add up to ≈1.5A, from Node analysis, the +IIN +-IOUT =Zero at any given node.
 
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