# homework problem

Discussion in 'Homework Help' started by kevinj, Dec 17, 2012.

1. ### kevinj Thread Starter New Member

Nov 24, 2012
4
0
Dont no how to get I1 or I2 and voltage drop only thing i can get is total resistance.

also is VS 2volts?

thanks

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2. ### tshuck Well-Known Member

Oct 18, 2012
3,527
679
Hello again,
Allow me to reiterate what I have previously stated:
If you've found the equivalent resistance, state it. I'm not going to take your word for it. At the very least, you can use it to check your answers.

Since you've asked about the voltage source, I'll assume Vs is the source voltage from V1 and V2....

No, Vs is not 2V. All that a voltage source does is say that the positive side is more positive than the negative side by X volts, where x is the number of volts for the supply. This being the case, V1 says that it's positive side, the longer bar on top, is 10 volts above common, or ground, since its negative side is connected directly to ground. V2 then says that its positive side is 8V above it's negative side, but this negative side is already 10V above common, since it's connected to the positive side of V1, so the voltage at the positive, with respect to ground/common, of V2 is.....(this is the part you have to do some work)

3. ### mrmount Active Member

Dec 5, 2007
59
7
If you have calculated the total resistance, you can find the total current
(I1) since the total voltage is given. I2 can be calcualted using the current divider rule after that.

4. ### tshuck Well-Known Member

Oct 18, 2012
3,527
679
The question, however, is asking for KVL and KCL solutions

We are working on that...

5. ### vk6zgo Active Member

Jul 21, 2012
677
85
Work it out anyway you can, ( if you haven't been asleep through all your classes,you should be able to do it "by inspection"),then do it again using KVL & KCL.

You should get the same results.

In the meantime, Google for "Series aiding" or similar,to understand about the battery voltages.

6. ### electronupdate New Member

Dec 20, 2012
2
0
1st calculate the equiv. series resistance of R2 and R3:

Requiv= R2 * R3 / (R2 + R3) == 2000 ohms

You can now calculate the current flowing through the whole circuit:

I = V / R
= (10 + 8) / (5K + 2K + 2.2k) [the 2K comes from above]
= 1.96 mA

Now, given that R2 and R3 have the exact same resistance they are going to share the current equally.

i.e. I(R2) = I(R3) = 1.96 mA/2
= 0.98 mA

Had R2 and R3 not been the same value, observe that the voltage across R2 and R3 must be the same.

This allows you two write:
V(R2) = V(R3) = I(R2)R2 = I(R3)R3

you can quickly see that this is ratio. Since you know also know the total current is the same as the current going through R1 you also know that I(R1) = I(R2) + I(R3). Two equations... and enough unknowns that one can grind it out...

7. ### WBahn Moderator

Mar 31, 2012
24,555
7,691
@electronupdate: Please note that this is the Homework Help forum, not the Homework Done For You forum. Keep in mind that, in general, problems posted here are problems that the student has been assigned, is expected to work on their own, and will receive a grade for. Also keep in mind that, again in general, the student has already seen example problems worked, either in the text or in class. So having one more problem worked for them by someone else is not going to move them much further along the road to comprehension. They need to struggle with putting things together for themselves. We can best help them by pointing out where they have gone wrong and giving just enough guidance to steer them toward the next step.