Resistor in paralell...

Thread Starter

ESPguitar

Joined May 15, 2005
16
Hi..

I have a relay the coil is 960 Ω and i need to couple a resistor in paralell with this relay, the total resistance should be 1547,32 Ω..

When i use the formula R2=(Rtot * R1) / (R1 - Rtot) i get this answer:

R2= (1547,32 * 960) / (960 - 1547,32) = 1455427,2 / -587,32 = (-2529,161615)

This ain't the right answer :S

Can somebody help me with this (i guess it is a easy question for you but, you know:S)

Thanks,

Robin
 

JoeJester

Joined Apr 26, 2005
4,259
You have a coil with 960 Ω resistance.

You have to couple a resistor to that coil to make the a total resistance of 1547.32

Am I correct?
 

Z2148

Joined Sep 21, 2005
1
Are you sure that Rtot=1547.32 Ω and R1=960 Ω ?

From what I know, Rtot should be smaller than the smallest resistance in the circuit when connected in parallel..
 

Dave

Joined Nov 17, 2003
6,970
Originally posted by Z2148@Sep 26 2005, 05:18 PM
Are you sure that Rtot=1547.32 Ω and R1=960 Ω ?

From what I know, Rtot should be smaller than the smallest resistance in the circuit when connected in parallel..
[post=10618]Quoted post[/post]​
Correct, you can't have the above mentioned situation for two resistive devices in parallel. Either the numbers are wrong or there is an additional resistive element in series with the parallel network.
 
unless you're using a faulty calculator that answer is impossible.

Rtotal = Rcoil*R / (Rcoil+R)

you can make R subject of formula. remember abs(Rcoil) = 2*3.142*freq*L

what's your inductance?
 

sanjeev203

Joined Sep 19, 2005
7
Originally posted by ESPguitar@Sep 25 2005, 09:40 PM
Hi..

I have a relay the coil is 960 Ω and i need to couple a resistor in paralell with this relay, the total resistance should be 1547,32 Ω..

When i use the formula R2=(Rtot * R1) / (R1 - Rtot) i get this answer:

R2= (1547,32 * 960) / (960 - 1547,32) = 1455427,2 / -587,32 = (-2529,161615)

This ain't the right answer :S

Can somebody help me with this (i guess it is a easy question for you but, you know:S)

Thanks,

Robin
[post=10589]Quoted post[/post]​
Robin,

In relay, resistor always goes in series with the inductor consituting (RL+R) as total resistance.

As other member suggested you, the equivalent resistance in parallel network is always lesser then the smallest one hence that has to be in series.

I found a nice article for you.

http://www.leachintl2.com/english/english2...erties/how8.htm

Hope this will help you.

Best regards,
Sanjeev
 
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