# Solving for E, R1, and I3 (picture included)

1. Jun 30, 2016

### Josh225

Sorry, I know I just posted a question but I picked up a textbook on circuit analysis over the summer just to teach myself some things and the examples during the chapter differ quiet a bit from the problems that they give you at the end of the chapter.

In the picture, it is asking to solve for E, R1, and I3.

I assume to find the voltage supply I would have to use E=Is (Rt) ... Would "Is" be 12.3 A since the sum of the current going in equals the sum of the current going out?

If it is... how would I find Rt if I don't know the value of R1. And how do you find the value of R1 if you do not know the value of Rt? Once I figure either one of the values out, I'd plug it into : 1/Rt= 1/R1 + 1/R2 + 1/R3

In my book I found a formula that said: I2 = (R1/ R1 + R2) It. When it was presented, that was only dealing with two resistors. Since here there are three resistors, would I thought it would be be : I3 = (R2/ R2+R3) It ? However, when I do that, I get the wrong answer...

The answers are supposed to be:
E= 36 V
R1= 24 ohms
I3 = 9 A

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2. Jun 30, 2016

### billy_joule

12.3A can't be used to find E, you'll need it to find R1 later.

Look at just the right hand half of the circuit, You know the resistance of both R2 and R3 and the total current flow. This is enough information to find E.

3. Jun 30, 2016

### Josh225

I'll give that a try once I get back home. Are the formulas I am using correct?

4. Jul 1, 2016

### Aristotle

This formula you speak of is called Current Division rule, useful when you want to find the current of a branch in a circuit... especially in this scenario--since current splits off in the resistors that are in parallel.
The 10.8 A current is the total current that flows out of R2 and that flows out of R3. Applying this to the formula you mentioned will get you the 9A!

Once you find your current, you will be able to use Ohm's law to find the voltage drop across the resistor

5. Jul 1, 2016

### Josh225

Thank you! I was under the impression that in order to find the voltage source, you needed to know Rt and It rather than just the values of R3 and I3. But now all is good. Thanks for clearing things up!

6. Jul 1, 2016

### Aristotle

The nice thing about resistors being in parallel nicely like that is that if you know the voltage drop across one of them, you'll know the voltage drop across the other resistors and the supply source--they share the same voltage
Just keep in mind current flow will differ across each resistor

7. Jul 1, 2016

### Josh225

Excuse my ignorance, but how would you find the voltage drop? I tried looking online, but there are alot of versions and im not sure which one i would use. You wouldnt use the voltage divider rule, would you?

8. Jul 1, 2016

### Staff: Mentor

I hope it actually said: I2 = (R1/(R1 + R2)) It.

That second set of parentheses IS NOT OPTIONAL, you know. A space on the right of the division sign does not equate to an invisible pair of parentheses there.

9. Jul 1, 2016

### Josh225

Hmmm, no it doesn't have that in there.

10. Jul 1, 2016

### Staff: Mentor

It almost certainly is there---unmistakably implied in the type-setting.

11. Jul 1, 2016

### Staff: Mentor

You apply Ohm's Law. Look for a known current flowing through a known resistance, and calculate their product.

12. Jul 1, 2016

### Aristotle

By using Ohm's Law of course!
Voltage drop across resistor (V) = Current flowing across a resistor (I) x Resistor's Value in Ohms (R)

13. Jul 2, 2016

### Josh225

When you multiply I3 x R3 you get 36. Wouldnt the voltage at R3 be lower than the voltage source itself because of the resistors?

14. Jul 3, 2016

### cnh1995

See how the resistors are connected. They are in parallel. They share equal voltage, which is the source voltage in this case.