# Resistor in paralell...

Discussion in 'Homework Help' started by ESPguitar, Sep 25, 2005.

1. ### ESPguitar Thread Starter Member

May 15, 2005
16
0
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

2. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,280
1,072
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?

3. ### Z2148 New Member

Sep 21, 2005
1
0
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..

4. ### Dave Retired Moderator

Nov 17, 2003
6,960
143
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.

5. ### aibelectronics Member

Aug 26, 2005
24
0
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

6. ### n9xv Senior Member

Jan 18, 2005
329
1
You must add a 580-Ohm resistor in series with the 960-Ohm coil. Assuming that 960-Ohms is the DC resistance.

7. ### sanjeev203 New Member

Sep 19, 2005
7
0
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