# Superposition Theorem Prob #2

Discussion in 'Homework Help' started by J.live, Oct 9, 2010.

1. ### J.live Thread Starter Member

Sep 10, 2010
35
0
a) Determine the contribution of the 1 A source to v1

b) Calculate the total current flowing through 7 ohms resistor.

Attempt:

a) Total resistor = (7+2) ll (5+5)= 4.737 ohms

v1' = iR = (1A)(4.737) = 4.737 ohms

• ###### Screen shot 2010-10-09 at 5.08.37 PM.png
File size:
19.6 KB
Views:
29
Last edited: Oct 9, 2010
2. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
For the $n^{th}$ time, watch your annotation! Voltage is expressed in Volts, not Ohms!

Your attempt at question a) is correct.

Since you have the voltage across the branch of the 7 and 2Ω resistors in series, you can calculate the current that flows trough them. That will be the first component of the current you look for.

Do the same analysis for the second source. This time the parallel branches will have 5, 7 and 2Ω on the left and 5Ω on the right. Find the second component of the current.

3. ### J.live Thread Starter Member

Sep 10, 2010
35
0
Looks like the book used the current division rule to find the current flowing through 7 ohms.

i= 4 x 5 / 5+7+2+5 = 1.05 A

V1"= iR = 1.05 x 9 = 9.475 V

V1= V1' + V1" = 14.21 V

i= V/R = 14.21 / 7 = 2.05 A

What I was trying to do was calculate the total resistance and then times that by 4. Pretty much the exact same procedure applied in part a. but that didn't work .

4. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
Are the equations above a copy from the book? The last line is wrong. V1 is applied across 7Ω+2Ω an thus I=V/R=14.21/9=1.579A.

Why don't you write down what you did and maybe we 'll sort it out.

Last edited: Oct 10, 2010
5. ### J.live Thread Starter Member

Sep 10, 2010
35
0
This is from an online solution manual for the book. Oh, thanks.