Finding current through a specific resistor

Thread Starter

jdole

Joined Sep 11, 2012
7
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
 

WBahn

Joined Mar 31, 2012
33,171
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
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?
 

Thread Starter

jdole

Joined Sep 11, 2012
7
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:

WBahn

Joined Mar 31, 2012
33,171
Is i = 0.2197 the right answer?
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.
 
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