Ohm's Law, Resistance, and Power

ΔV=I*R
R=ΔV/I
Power=I*ΔV

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

Simon Bridge
Homework Helper
I said that ΔV of A and ΔV of B are the same (each 3/2 V)
The resistance of B and C in parallel is less than the resistance of A, so the volt-drop will be less.

How so? They're each 10 ohms.

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.

CWatters
Homework Helper
Gold Member
How so? They're each 10 ohms.

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.

Simon Bridge