# Parallel Resistors and Current HW (pics)

Discussion in 'Homework Help' started by breck, Sep 17, 2011.

1. ### breck Thread Starter New Member

Sep 17, 2011
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0
3 Problems I am having on my HW (picures below)

I have read through the DC chapter which this is all based off of, but there these problems, I can't seem to figure out how to write the correct formulas to solve them.

3. I assume some type of equation relating the resistors to each other will be used.
R_tot=R1+R2+...
or
R_para=R1R2/(R1+R2)

That is basically where I am on on the first one. I don't know how to relate R_L to the R_AB to get R.

5. Some derivation of V=IR where R=R_para

Not sure how to get around it.

6. I have no idea how to approach this one. There are no examples of this type of problem in my book and I honestly don't know what to search on the internet to try to find something like this.
I have absolutely no clue on how to get to the answer, even though my answer was the opposite of the answer, it was a percentage guess from my previous answer and I guess because R1R2-R3R4 is negative it leads to the resultant being negative but I am not sure what so ever.

Any help would be lovely. Thank-you all.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
For question 3 you have Rab = RL= 515Ω
So
Rab = R + R||(R+RL)and you solve for R

Question 5
http://en.wikipedia.org/wiki/Current_divider
Ix = Itot * R2/(R2+R3)

Itot = V / (R1 + R2||R3)

Q6
You can use voltage divider rule

Vx = (V*R2)/ (R1+R2 ) - (V*R4)/(R4+R3)

Last edited: Sep 17, 2011
breck likes this.
3. ### breck Thread Starter New Member

Sep 17, 2011
3
0
Thank you for the help, but my math may be a little bit wrong for #3.
515=R+(515R)/(515+R)
515=((515+R)R+515R)/(515+R)
515(515+R)=R^2+1030R
0=R^2+515R-265225
(-515+/-sqrt(515^2-4(-265225)))/2
R=318 and -833
R is supposed to be 297.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Well I made the mistake in the equation
Rab = R + R||(R+RL) this looks good
So we have
515 = R + (R*(R + 515))/(R + (R + 515))

R = 515/√3 = 297.335Ω

Sorry for the mistake

breck likes this.
5. ### breck Thread Starter New Member

Sep 17, 2011
3
0
No worries. Thank you for taking the time to help me out.