Resistors in a Circuit

1. Jan 13, 2010

the-alchemist

1. The problem statement, all variables and given/known data
I have been stuck at the following question for some time now. All help is appreciated.

2. Relevant equations
P=IV, V=IR

3. The attempt at a solution
I have resolved the right side of the circuit into a 10 ohm resistor. I have also found the current in the 20 ohm resistor (0.4A). I then used Ohm's Law to get the current in the 10 ohm resolved resistor and found it to be 5A. Then using Kirchoff's Current Law, I got 5.4A in the unknown resistor R, which gives a resistance of 9.26 ohms.

This 9.26 ohms however disagrees with the given answer of 10 ohms.

Where have I went wrong? All help is appreciated!

Thanks!

2. Jan 13, 2010

vela

Staff Emeritus
You didn't apply Ohm's Law correctly. 50 volts isn't the voltage drop across any of the individual resistors; it's the voltage drop across the entire combination of resistors. Each resistor will have a smaller voltage drop across it.

If you take the two equations you have, you can combine them to get

$$P=\frac{V^2}{R}$$

P is the power dissipated by a resistor of resistance R, and V is the voltage drop across the resistor. Since you're given the power dissipated by the 20-ohm resistor, you can use the formula to calculate the voltage across the 20-ohm resistor. Then you can use Ohm's Law to calculate the current through the resistor.

Ultimately, what you want to figure out is the voltage drop across the unknown resistor and the current flowing through it because then you can use Ohm's Law to calculate its resistance.

3. Jan 13, 2010

the-alchemist

I've tried this method as well but I get stuck after calculating the voltage for the 20 ohm resistor. Correct me if I'm wrong but I thought any resistor that is parallel to an ideal voltage source has no effect on the voltage or current in the rest of the circuit.

I'm still unable to get the answer of 10 ohms. Shucks.

4. Jan 13, 2010

vela

Staff Emeritus
Yes, this is true. The voltage source will source as much current as is needed to maintain a fixed voltage across its terminals. Anything connected to those terminals will see that voltage drop regardless of what else is connected to them as well, like a resistor.

Two elements in a circuit are in parallel only if they are connected to the same two nodes of the circuit, so in the circuit, none of the resistors is in parallel with the voltage source because none of them connect directly to both the top and the bottom of the source.

What did you get for the current and voltage for the 20-ohm resistor?

The 20-ohm resistor and 10-ohm resistor are in parallel, right? So what quantity do they have in common?

How is the current through R related to the currents through the 20-ohm and 10-ohm resistors?

5. Jan 13, 2010

the-alchemist

This really helped me! Thanks!

I think I got it now. Correct me if I'm wrong, the current for the 20 ohm resistor is 1A and voltage of 20V.

The 20 ohm and 10 ohm resistors are parallel thus they share the same voltage drop.

Using the voltage divider principle I was able to get 10 ohms for R after resolving the parallel resistors 20 and 10 ohms into 6.67 ohms with 20V as the voltage drop across the 6.67 ohm resistor. Can anyone confirm that this method/ way of doing is appropriate and correct?

6. Jan 13, 2010

vela

Staff Emeritus
Yes, that's correct.

You could have also calculated the current through the 10-ohm resistor since you know the voltage across it. The current through R would be the sum of the currents through the 10-ohm and 20-ohm resistors, and voltage across R would be the difference between the 50 V supplied by the battery and the 20-V drop across the 20-ohm and 10-ohm resistor combination. Divide the voltage by the current, and you'd get the same answer.