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Superposition problem 2

  • #1
Find the total amount of current flowing through R3 in the circuit below.

I mainly have doubts whether the answer given in the book is correct, to me it doesnt seem correct. Heres my attempt anyway.

http://img409.imageshack.us/img409/7584/superposition2iu4.png [Broken]

Make VS1 a short.

Rt = R1//(R2//(R3+R4)
(R1 is in parallel with R2 which is in parallel with the series combination of R3 and R4).

Rt = 680//(220//(330+470) = 680//(220//800) = 680//172.54 = 138.89ohms

Total current flowing from current source = 100mA

IR3 = Rt/(R3+R4) * IS = 139/800 * 100mA = 17mA

Current is flowing left to right over R3.
-------------------------------------------------------------------------

Make IS an open as it is a current source.

Rt = R2 + ((R3+R4)//R1)
(R3 and R4 are in series, this combination is in parallel with R1, R2 is in series with the rest of the circuit.)

Rt = 220 + (800//680) = 587

It = Vs1/Rt = 20/587 = 0.034A or 34mA

IR3 = Rt / (R3+R4) * It = 587/800 * 0.034 = 0.025A or 25mA
(i chose R3+R4 as the current flowing through a series combination is the same.)

As the polarity of both sources is the same (current flowing in same directions). The total current across R3 = 42mA

If there are any errors could someone please indicate them to me, any help would be very much appreciated.
 
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Answers and Replies

  • #2
the back of the book says 1.6mA is flowing through R3, to me doesnt seem to make any sense.
 
Last edited:
  • #3
berkeman
Mentor
56,814
6,782
A good way to cross-check your answer would be to write the two KCL equations for the two sides of R3, and solve for the voltages and the resulting R3 current. What answer do you get when you solve this with straight KCL equations?
 
  • #4
SGT
Find the total amount of current flowing through R3 in the circuit below.

I mainly have doubts whether the answer given in the book is correct, to me it doesnt seem correct. Heres my attempt anyway.

http://img409.imageshack.us/img409/7584/superposition2iu4.png [Broken]

Make VS1 a short.

Rt = R1//(R2//(R3+R4)
(R1 is in parallel with R2 which is in parallel with the series combination of R3 and R4).

Rt = 680//(220//(330+470) = 680//(220//800) = 680//172.54 = 138.89ohms

Total current flowing from current source = 100mA

IR3 = Rt/(R3+R4) * IS = 139/800 * 100mA = 17mA

Current is flowing left to right over R3.
-------------------------------------------------------------------------

Make IS an open as it is a current source.

Rt = R2 + ((R3+R4)//R1)
(R3 and R4 are in series, this combination is in parallel with R1, R2 is in series with the rest of the circuit.)

Rt = 220 + (800//680) = 587

It = Vs1/Rt = 20/587 = 0.034A or 34mA

IR3 = Rt / (R3+R4) * It = 587/800 * 0.034 = 0.025A or 25mA
(i chose R3+R4 as the current flowing through a series combination is the same.)

As the polarity of both sources is the same (current flowing in same directions). The total current across R3 = 42mA
If there are any errors could someone please indicate them to me, any help would be very much appreciated.
No, the polarity of Vs1 makes the current in R3 negative, so you must subtract the currents.
The rest of your reasoning is correct, so if did not make any mistake in the calculations, the answer should be -8mA.
 
Last edited by a moderator:
  • #5
So am I right in assuming the 1.6mA is not the correct answer? Im fairly sure that my calculations are correct, I will have another look. Thank you SGT for your pointers on series parallel relationships, the final cog turned this morning! Does anyone have any recommended texts that mainly focus on problems and worked solutions? I have lots of theory books but not enough problems to work my way through.
 

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