missing resistor

Thread Starter

phantomvs

Joined Jan 19, 2016
39
I am stuck on another question that asks me to find the missing resistor of AB in the attached diagram.
I know the following :
R1= 30 ohms
R2= 40 ohms
R3 = 10 ohms
R4= 10 ohms
R5=60 ohms

Now i know the formulas for calculating the R value for parallel and series circuits but can't figure it ou.
I tryed seperating the circuits in smaller circuits .
I did R2 & R5 as parallel and do i do series as R1 , R3 and R4 ?

Thanks !
 

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WBahn

Joined Mar 31, 2012
25,897
This makes no sense. Think about it. If I connect up those five resistors according to the diagram provided, how could you possibly tell that there even IS a resistor that is missing between A and B, let alone what value it is?

You are missing some piece of information that gives the resistance between some pair of points in the circuit that involves the missing resistor.
 

Thread Starter

phantomvs

Joined Jan 19, 2016
39
This makes no sense. Think about it. If I connect up those five resistors according to the diagram provided, how could you possibly tell that there even IS a resistor that is missing between A and B, let alone what value it is?

You are missing some piece of information that gives the resistance between some pair of points in the circuit that involves the missing resistor.
Your right, but that is the only info i see. I found it wierd that there is no voltage or amps indicated anywhere in the circuit.
 

WBahn

Joined Mar 31, 2012
25,897
They are not asking for the value of some missing resistor between A and B. They are asking for the equivalent resistance of THOSE resistors as seen between A and B.
 

Thread Starter

phantomvs

Joined Jan 19, 2016
39
I think i figured this one out thanks to ABSF new diagram.

I found the total resitance of R2, R5 and R1 which is 13.33 ohms

Then added in R3 and R4 for 33.33 Ohms

Just re-drawing the schematic, i don't get why R1 goes in parallel when its in series in original diagram.
Cause R2 and R5 are in parallel which in turn is in R1.
 

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WBahn

Joined Mar 31, 2012
25,897
I think i figured this one out thanks to ABSF new diagram.

I found the total resitance of R2, R5 and R1 which is 13.33 ohms

Then added in R3 and R4 for 33.33 Ohms

Just re-drawing the schematic, i don't get why R1 goes in parallel when its in series in original diagram.
Cause R2 and R5 are in parallel which in turn is in R1.
Two resistors are in series if they are the only two resistors on a given node.

Two resistors are in parallel if both ends are connected to the same nodes.

Identify the nodes.

A simply way to do that is to color them.

colored.png
R1 is connected to the red node and the greed node, thus it is in parallel with any other resistor connected to the red node and the green node.

The resistors on the green node are R1, R2, R3, and R5. Of these, R1, R2, and R5 are also connected to the red node. Thus R1, R2, and R5 are all in parallel.
 

Thread Starter

phantomvs

Joined Jan 19, 2016
39
You see thats my problem its hard for me to tell if a resistor is in series or parallel.

The original diagram , it looks like R1 is in series.
 

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WBahn

Joined Mar 31, 2012
25,897
Remember the definition of series. To be in series, whatever current flows in one MUST flow in the other. That means that whatever current flows out of one has to have no choice but to enter the other. Look at that green node and imagine a current flowing up out of R1. Does it HAVE to flow into R2? No, there is another path it could take. Similarly, consider current coming up out of the supply. Does the current then goes through R1 have to also go through R2? No.

In order for the current flowing out of one to have to flow into the other, there must be no other paths available. If the two resistors are connected directly together at a certain node, this means that there must be no other devices connected to that same node.
 
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