I thought i was doing this right =( what am I doing wrong?!?

Question

In a series circuit, R1=8 Ohms, R2=12 Ohms, and R3=16 Ohms.

I need to find the voltage drop on resistor R2!

I found the equivalent resistance to be 36 Ohms.

I used the equation V=I*R

I=V/R=12V/36 Ohms=v/3 Ohms

Then I substituted this into the equation..

V=1/3 * 36 Ohms

V=12V

What am I doing wrong?

Please Help Me!

Answer 1

You got your units mixed up

I = V/R = 12/36 = 1/3 Ampere (the unit for current)

V2 = I*R2 = (1/3)*12 = 4 Volts

Answer 2

For one where did the 12 Volts come from when you did the I = V/R stuff? I will assume that you mean the voltage drop across all three resistors is 12 Volts.

Well your mistake might be in the line after you say "Then I substituted this into the equation".

You want the voltage drop across the second resistor for whom R = 12 Ohms and not 36 ohms. You just found the voltage drop across all three resistors.

So V = IR = (1/3)12 = 4 Volts

Answer 3

OK ... let's do it step by step.

1. Take the equivalent resistance of the three resistors. Since they are in series, the equivalent resistance is

Re = 8 + 12 + 16 = 36 ohms

2. Note that the current in the circuit is CONSTANT. Hence, using Ohm's law,

V = IR

where

V =voltage

I = current

R = resistance

and solving for "I"

I = V/R = 12/36 = 1/3 amp

3. Since the current in this circuit is constant, then the voltage drop in resistor 2 is

V(2) = (1/3)(12) = 4 volts

Hope this helps.