# Ohm's Law, Resistance, and Power

1. Feb 28, 2013

### jperentosh

1. The problem statement, all variables and given/known data

2. Relevant equations

ΔV=I*R
R=ΔV/I
Power=I*ΔV

3. The attempt at a solution

I said that ΔV of A and ΔV of B are the same (each 3/2 V). Using this, I got that the current at A = 3/20 Amps. However, this means that the current at B and C is each 3/40 Amps. However, then the 1.5 V * 3/40 does not equal 10 ohms, which is given. How am I supposed to go about this problem to find the power dissipated in each bulb and find what changes in terms of bulb brightnesses in A and B if bulb C is unscrewed?

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2. Feb 28, 2013

### SteamKing

Staff Emeritus
You have one resistor in series with two others which are in parallel. You have not found the correct current in the circuit.

3. Feb 28, 2013

### Simon Bridge

The resistance of B and C in parallel is less than the resistance of A, so the volt-drop will be less.

4. Mar 1, 2013

### jperentosh

How so? They're each 10 ohms.

5. Mar 1, 2013

### resigned

J, they are in parallel. There are 2 possible paths for current so there is less total resistance for electrons being pushed through this path. You can think of this sort of like having 2 checkout lanes open at a grocery store. More shoppers can move through faster than if there were one lane open.

Resistors in series (in a row) are added as R1+R2. Resistors in parallel (multiple paths) are not added this way. Hint: If there are two parallel paths, both must have the same voltage across them.

6. Mar 1, 2013

### CWatters

Apply KCL to the point where A is connected to B and C. You will find the current through B (or C) will be half the current through A. So the voltage drop oved B will be half that of A.

7. Mar 1, 2013

### Simon Bridge

Thanks folks - jperentosh: yes they are correct. By now you should have seen the equations for resistors in series and parallel. You can always check with an ohmmeter if you don't believe it.