# Resistance Problem Verification

Discussion in 'Homework Help' started by EL7819, Dec 7, 2011.

1. ### EL7819 Thread Starter New Member

Apr 15, 2011
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A 24-volt relay coil draws 48mA at rated voltage. A resistance check showed the coil to have 100ohms resistance. How much resistance should be added to the circuit in order to operate the relay coil at 120 volts?

Here is my solution:

$24V/48mA= 500ohms =Z$

$R=100ohms$

$R^2+X^2=Z^2$

$sqrt(100^2-500^2)=X$

$sqrt(240000)=489ohms$

$120V/48mA=2500ohms$

$Z=2500ohms
X=489ohms$

$sqrt(Z^2-X^2=R$

$sqrt(6250000-239121)=2451ohms=R$

$2451ohms-100ohms(already in circuit)=2351.71ohms$

Hello,
I wanted to see If I have the correct answer.
Thank you

Last edited: Dec 8, 2011
2. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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Don't do it. Get a 120VAC rated relay.

3. ### bountyhunter Well-Known Member

Sep 7, 2009
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Check again. The coil resistance is 500 Ohms if it draws 48 mA with 24V across it.

Dec 26, 2010
2,147
298
Not necessarily. 120V being a possible mains voltage in some countries, the relay may be an AC one, when the coil inductance would be significant. The OP seems to have been using Pythagoras' theorem, which rather points in that direction.

EL7819 likes this.

Dec 26, 2010
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@EL7819: To check your answer, use the numbers you have calculated to calculate the total impedance, and check how much current will pass when 120V is applied.

Is it reasonably near 100mA? How close would it be if you chose a resistance that somebody could actually buy?
(Your course may not require this, but it would be necessary in a real engineering situation. )

Note that you seem to have written a subtraction the wrong way around: as shown you would require to find the square root of a negative number.

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6. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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This is in the homework section, so I suppose he may be trying to solve an academic problem. But it is a bad thing to do.

If the coil is designed for 48VAC, it is using magnet wire and geometry for that voltage. Even if a resistor is used to drop the voltage, an instantaneous peak of 170V can still present itself to the coil at turn-on (current at t0 is 0, so the voltage across the resistor is 0), which can cause insulation break down. Eventually, this might result in smoke.

Dec 26, 2010
2,147
298
This almost certainly is a set exercise. While one should never risk over voltage on anything, it may be a bit surprising to find a relay coil designed with its insulation unable to tolerate considerable short-term excess, as very large transients can result whenever the supply to a relay coil is switched off. Whether or not a relay is used with a dropping resistor, better practice might have something like a bidirectional TVS across the coil to suppress over-voltage transients.

It seems likely however that the main interest of the original question was to solve impedance triangles, rather than actually to design a relay circuit.

8. ### EL7819 Thread Starter New Member

Apr 15, 2011
20
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$24V/.048=500ohms=Z$

$X_l=100ohms$

$R^2+X^2=Z^2$

$sqrtZ^2-X^2=R$

$sqrt500ohms^2-100ohms^2=R$

$sqrt240000=489ohms=R$

$120V/.048=2500ohms=Z$

$sqrtZ^2-R^2=X$

$sqrt6250000-239121=X$

$X=2451.70ohms-100ohms=2351.70ohms$

$sqrtR^2+X^2=Z$

$sqrt6007401+239121=2499.99=2500ohms=Z$

9. ### EL7819 Thread Starter New Member

Apr 15, 2011
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Last edited: Dec 8, 2011
10. ### bountyhunter Well-Known Member

Sep 7, 2009
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He said it had 24V on it, did not specify AC voltage or any frequency.

Going with what he posted.

Dec 26, 2010
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298

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Dec 26, 2010
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If the original statement was a set question, with the given values of voltage, current, and resistance stated and 120V input specified,
I would think an AC input most likely unless the question has been copied mistakenly.

Possibly your advice has confused the OP, or possibly mine has. He will find out what is right when he submits his answer.

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13. ### EL7819 Thread Starter New Member

Apr 15, 2011
20
0
A 24-volt relay coil draws 48mA at rated voltage. A resistance check showed the coil to have 100ohms resistance. How much resistance should be added to the circuit in order to operate the relay coil at 120 volts?

Not going with what he posted but just did the extra work to prove a point that it wouldn't make any difference calculation wise. The original way I did the problem I feel more confident by and it makes more sense to me.

I think this is AC Relay.

Concerning the problem, Do I check this by taking the new voltage(120V) by the calculated impedance(2500ohms)?

Last edited: Dec 8, 2011

Dec 26, 2010
2,147
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Yes, calculate the impedance from the inductive reactance, and the sum of the coil resistance and the added resistance.

Divide the supply voltage by the impedance.

Z= √R$^{2}$ + X$^{2}$), I = V/Z

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15. ### EL7819 Thread Starter New Member

Apr 15, 2011
20
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$Z=2500ohms
X=489ohms$

$Z=2500ohms
X=489ohms$

$sqrt(2451^2ohms+489^2johms=2499.30=2500ohms$

$120V/2500ohms=.048mA$

Thank you.

Now just for sake of discussion, most of the relay resistances I found had not too much resistance. In a real world situation, If a customer wanted something like this done would the danger behind it be adding the other relay in with the other one?

Dec 26, 2010
2,147
298
$

I really do not understand what you are trying to ask: could you put that another way?

"...most of the relay resistances you found had not too much resistance..." (Too much resistance for what, exactly?)

"...would the danger behind it be adding the other relay in with the other one?"
(The danger behind what? Are you planning to use two relays, if so why? This was not mentioned until now.)$

17. ### EL7819 Thread Starter New Member

Apr 15, 2011
20
0

Last edited: Dec 9, 2011

Dec 26, 2010
2,147
298
Resistors are most often made in standard values. Possibly the commonest values nowadays are those of E24 series, which approximates to a geometric progression of 24 values for every ten times increase of resistance (that is, 24 values per decade).

The E24 series is typically used with 5% resistors (closer tolerance resistors are sometimes made in these values, as well as in specialised series like E48, E96...). There may not be much need to set the voltage dropper for a relay more accurately than you could get it with one E24 resistor. Excluding tolerance, this should get you within about 5% of your target, if you choose the nearest value up or down. Some people might think E12 good enough, but that would double the error.

You might choose to use two resistors of different values in series or parallel if you did want to get closer, or if only E12 or coarser ranges were available. There are calculators for that available on line, I think (I vaguely remember something on this site, somewhere), or you might want to make up a spreadsheet for it yourself an exercise.

http://en.wikipedia.org/wiki/Preferred_number#E_series

http://www.logwell.com/tech/components/resistor_values.html

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