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ritamonteiro ritamonteiro
wrote...
Posts: 25
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11 years ago
Three 9.0 ohm resistors are connected in series with a 12 V battery.

Find the current in each resistor.

I know the current should be the same for all the resistors but i'm not sure i'm doing it right.
I have 0.44 A. Is this right? any helpful suggestions?
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3 Replies
Replies
wrote...
11 years ago
Alright, you already know total voltage (12V). In a series circuit, total resistance is the sum of all of the resistors, therefore, because there are three 9.0ohm resistors, total resistance is equal to 27 ohms.

Current = total voltage / total resistance
Current = 12V / 27ohms
Current = 0.44A

So, in a nutshell, yes. You are right.
Answer accepted by topic starter
ijhsjijhsj
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11 years ago
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wrote...
11 years ago
Resistors in series can be written as one big equivalent resistor, with resistance equal to the sum of each of them.

So in this case the equivalent resistance would be 27.0 ohms (3 x 9.0 ohms).

Using Ohm's law, I = V/R, you discover that the current through the circuit is 0.44 A, so you're right.

Remember that for a simple series circuit, the current through the entire loop is everywhere equal due to Kirchhoff's Current Law (KCL). (The net current going into any node must be zero.)

This means if you pick any point in the circuit, you can draw one current going into it from one side (I1) and another going into it from the other side (I2) then using KCL, you can write

I1 + I2 = 0
or
I1 = -I2

This negative sign looks strange at first, but keep in mind that you drew both arrows going INTO the node, meaning the value of I2 itself will be negative (similar to setting up an arbitrary coordinate system in a mechanics problem).

This basically means that along any branch, current into any point = current out of that point. So the current going through each resistor - and the voltage source - must be the same.
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