Parallel circuits

Thread Starter


Joined Feb 5, 2008
Hi everyone i'm new to the forms.
I'm taking a fundamentals of electricity class (not by choice )and I’m completely lost. Right now we are doing parallel circuits and I’m stuck on how to figure out the voltage drop across each resistor, the power dissipated by each resistor and the power supplied by the battery. I would like to say that I’m complete dumba** when comes to this stuff. Any help would be great, thanks in advance. :confused::confused::confused:


Joined Jan 28, 2005
Can you post a sketch of your circuit to help us in assisting you?


Thread Starter


Joined Feb 5, 2008
I tried adding an attachment last night after i posted this, but i didn't see where to add an attachment.


Joined Feb 6, 2008
1. go to somewhere like or
2. create an account
3. upload the photo
4. put the url of the image in between img tags.
5. finished


Joined Nov 17, 2003
We do support uploading of attachments.

I tried adding an attachment last night after i posted this, but i didn't see where to add an attachment.
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Joined Jan 25, 2008
To start lets think about the conceptual first. Take for example Christmas lights. What happens when one bulb burns out/ is taken out? A part of the string turns off while the rest of the string works. This is because the Christmas lights are wired both in parallel and in series. the bulbs that go out are in series with each other and as such once you remove a bulb you have an open circuit in that part. Basics of electricity are that an open circuit has about an inf resistance and thus a zero current (no electricity flows through it).

Now with parallel, voltage will be content along the parts in parallel; thus the sections of lights in parallel remain on even with part of the string open. Another conceptual aspect of parallel is consider 10 light bulbs attached to a battery. If they are all in series they will be dim; but if they are in parallel they will all be bright.

Conceptual stuff out of the way we will be working with two main equations. V=IR and P=IV. Now if we have a circuit like what is attached we will see we have R1 in Parallel with the sum of R2 and R3.
Because they are in parallel we know the voltage drop across both lines are the same (12V) Because I=V/R, I2 will equal 12V/R1 while I3 will equal 12V/(R2+R3). At the same time I1=I2+I3 because we cannot magically create current out of thin air. Thus this is a current divider because I1 is being divided into I2 and I3.
Now lets take a look at just R2 and R3. We have I3 going across both of them and this current will not change because they are all on the same line. Knowing that I3 stays the same and that V=IR we know that the voltage drop across R2 is I3*R2 and the voltage drop across R3 is I3*R3. Also note that the sum of the voltage drops is 12; thus this is a voltage divider.

As far as Power is concerned you are looking at two of the three aspects at any given location. remember that P=IV=(I^2)R=(V^2)/R. So the power supplied by a battery will be the voltage of the battery times the current it is producing or the voltage of the battery squared divided by the total resistance. The power at any resister will be the current squared times the resister or the voltage squared divided by the resister.

Hope this helps.