# Finding current through a specific resistor

Discussion in 'Homework Help' started by jdole, Sep 19, 2012.

1. ### jdole Thread Starter New Member

Sep 11, 2012
7
0
A question on my exam asked me to find the current through a 7.5Ω resistor. I simplified the circuit to this:

http://i.imgur.com/5hVR2.png

I was wondering if it acceptable to add up the voltages and resistors to find the current because that's what I did. I thought it was fine because it's a series circuit.

I got 7V for the equivalent voltage and $\frac{355}{26}$Ω for the equivalent resistance.

The current I got was 0.5127 A

2. ### mlog Member

Feb 11, 2012
276
36
The current should be negative (clockwise) based on the direction of the diagram.

3. ### jdole Thread Starter New Member

Sep 11, 2012
7
0
Last edited: Sep 19, 2012
4. ### WBahn Moderator

Mar 31, 2012
24,555
7,691
One of the nice thing about most circuit problems is that you can usually check the answer your self.

If you have 0.5127A flowing through the resistors, what is the voltage drop across each? How does the sum of these compare to the sum of the algebraic sum of the voltage sources? (and remember, +7V and -7V are not the same!)

After reversing the 5V source, what results do you get when performing this check?

5. ### jdole Thread Starter New Member

Sep 11, 2012
7
0
Dang, I was rushing because I had 1 minute to turn in my exam, so I added 2 to 5 instead of subtracting 2 from 5.

I got 0.2197 as the current.

Hopefully I get partial credit.

Last edited: Sep 19, 2012
6. ### jdole Thread Starter New Member

Sep 11, 2012
7
0
Is i = 0.2197 the right answer?

7. ### WBahn Moderator

Mar 31, 2012
24,555
7,691
Does it pass your checks?

(Yes, it is correct)

If you showed your work with enough detail so that the grader can see what you got right and where you went wrong, then they have a basis for assigning partial credit (whether they do or not is, of course, up to whatever their policy is). But if you just had some scribbles and an answer, then there is little basis for partial credit.