Simple resistance question on D.C. Circuits

[Punch]

Homework Statement

Find the resistance of L1

The Attempt at a Solution

R= V / I
= 12 / 0.5A
=24 ohm

However, the answer stated is 16 ohm... Couldn't spot any error in my workings, can someone kindly show me the right way out? :(

 

[kuruman]

It is true that 0.5A is the current through the lamp, but it is not true that the voltage across it is 12V. That would be the case if the three resistors were not there. Use the fact that the three resistors are identical to find how much current flows in each branch. Knowing this and that the voltage across R3 is 2V, find what the resistance of the three resistors is. Solve the circuit from there.

 

[thebigstar25]

the 24 ohm you found is not for the lamp.. imagine that instead of L1 they gave you R5, work it out this way and you will end up with the right answer..

 

[Punch]

Then, would I be right to say that R1=R2=R3=2V?

The more I think about it, I can't make any connection still... sorry to trouble you but may I request for an explanation or working please?

 

[thebigstar25]

Then, would I be right to say that R1=R2=R3=2V?

The more I think about it, I can't make any connection still... sorry to trouble you but may I request for an explanation or working please?

we can not solve the problem for you .. all we can do is giving you few hints..

from the question, it said that you have three identical resistors R1, R2, and R3 .. you can say that all of them have a resistance of R ..

note that R2 and R3 are connected in series, and their equivalence is connecting in parallel with R1, apply your knowledge of series and parallel connections regarding the current and the voltage, write as many equations as you can get, and try to think how you can extract from them the right answer..

 

[OmCheeto]

Then, would I be right to say that R1=R2=R3=2V?

Nope.

The more I think about it, I can't make any connection still... sorry to trouble you but may I request for an explanation or working please?

You know that the voltage across R3 is 2 volts. Since R3 and R2 are in series, the current flow through them will be the same. And since the resistances are all identical, from Ohm's law you can deduce that the voltage drop across R2 will also be 2 volts.

But the voltage drop across R1 will not be 2 volts.

Listen to our friend Gustav to get the answer:

the total voltage around a closed loop must be zero.

 

[Punch]

How about this? R1 is parallel to R2 and R3.

So by the parallel circuit rule, V=V1=V2

Hence, R1=R1+R2
=2V + 2V
=4V

Then voltage running around the circuit is 4V excluding L1?

Hence by the series rule, 12V=L1+4V
L1=8V

 

[thebigstar25]

How about this? R1 is parallel to R2 and R3.

So by the parallel circuit rule, V=V1=V2

Hence, R1=R1+R2
=2V + 2V
=4V

Then voltage running around the circuit is 4V excluding L1?

Hence by the series rule, 12V=L1+4V
L1=8V

excellent! now you figured out that the voltage is 8V for L1, you want the resistance of L1, and in order to find out this, you need voltage and current so you can apply ohm`s law .. you already found out what is the voltage, can you find out what should be the current?

 

[Punch]

excellent! now you figured out that the voltage is 8V for L1, you want the resistance of L1, and in order to find out this, you need voltage and current so you can apply ohm`s law .. you already found out what is the voltage, can you find out what should be the current?

8V=0.5A * R
8 / 0.5 = R
R=16 ohm ^_^

 

[thebigstar25]

see you got it yourself, next time don't give up so fast :)

 

[Punch]

Thanks ^^ I figured that out by applying parallel circuit's rule, V=V1=V2 and the fact that the resistor were identical. I couldn't solve the problem because I was confused by thinking that R1 was also 2V.

However, I have since acknowledged that this wasn't true... But.. I am still a little confused... why this is so

 

[thebigstar25]

ok , let us start first from what we know ..

http://img203.imageshack.us/img203/7976/58147717.jpg


we know that if we have the situation illustrated in the previous figure that the voltage across R1 and R2 is the same (even if R1 not equal R2) .. I think we agree on that ..


now let's go back to the figure in your question, which is as follows:

http://img690.imageshack.us/img690/5724/96169558.jpg


if you compare your question to the figure I put, you will find that it would be almost the same situation if we had one resistor instead of the two resistors R2 and R3 (imagine that there is no ameter or L1) .. so at this point in order to be able to deal with this circuit, we have to convert it somehow to something we already know, which in our case we will try to make the circuit looks like the first figure in this post ..

I have done the job, notice what I did in the following figure:

http://img404.imageshack.us/img404/5602/72951149.jpg


in the process to get from (1) to (2), I put R2 and R3 in a box and I called that box Req .. and as you can see, (2) is similar to the first figure in this post .. and here you can say that the voltage across R1 is the same as Req .. and in our case Req is just R2+R3 (since they connected in series) ..


I hope you got it now.. if you are still confused about something please ask again.. :)

 

[Punch]

Wow, I think I just understood it completely!

Thanks, so 12V=voltage of L1 + R1

or 12V=voltage of L1 + (R2+R3 voltage)

 

[thebigstar25]

exactly :) ..