Voltage Divider Rule on AC Circuit

[leejersoon]

Hi everyone,

I'm not familiar of AC circuits as compared to DC circuits. But someone mentioned to me that I can use the Voltage Divider Rule (VDR) on AC circuits to drop a 240 Vac to 24 Vac by using a high watt 1 ohm resistor and a 9 ohm resistor.

(240*1) / (9+1) = 24 Vac.

I was thinking about it and it kind of make sense. But won't it be up-side-down when the AC current flows from the other way? Then it will become:

(240*9) / (1+9) = 216 Vac.

This will cause the AC voltage to be unstable....right?
Which means for this to work I can only use the resistors that are the same value?

(240 * 5) / (5 + 5) = 120 Vac.

Am I right or did I go wrong some where or is this not applicable at all? :? Thanks.

 

[charlie b]

But won't it be up-side-down when the AC current flows from the other way? Then it will become:
(240*9) / (1+9) = 216 Vac.

No, it works the same in both halves of the cycle. The voltage of an AC circuit varies moment-to-moment. At one point, it will be lower than 240 at the source, and it will reach a peak value that is higher than 240. The value of 240 is a "sort of average" value. But at any given moment, the voltage across your 1 ohm resistor will be one tenth the value of the source. If the source is at +136 (i.e., during the positive half cycle) at a given moment, the voltage across the 1 ohm resistor will be 13.6 at that same moment. If the source is at -200 at a given moment (i.e., during the negative half cycle), the voltage across the 1 ohm resistor will be -20 at that same moment.

 

[Pharon]

The value of 240 is a "sort of average" value.

Or more specifically, the Root Mean Square (RMS) value -- which is (peak voltage) * (sqrt(2)/2), I believe.

 

[ggunn]

Hi everyone,

I'm not familiar of AC circuits as compared to DC circuits. But someone mentioned to me that I can use the Voltage Divider Rule (VDR) on AC circuits to drop a 240 Vac to 24 Vac by using a high watt 1 ohm resistor and a 9 ohm resistor.

(240*1) / (9+1) = 24 Vac.

I was thinking about it and it kind of make sense. But won't it be up-side-down when the AC current flows from the other way? Then it will become:

(240*9) / (1+9) = 216 Vac.

This will cause the AC voltage to be unstable....right?
Which means for this to work I can only use the resistors that are the same value?

(240 * 5) / (5 + 5) = 120 Vac.

Am I right or did I go wrong some where or is this not applicable at all? :? Thanks.

One thing you need to consider with any voltage divider is that the load resistance also contributes to the network and therefore influences the voltage at the center node. Also, a ten ohm resistance across 240V is going to draw 24A and dissipate 5760 Watts. That's going to produce a lot of heat even if your resistors can take it.

 

[drktmplr12]

The problem lies in, as ggunn pointed out, when you connect a load to this resistor network, the dynamic changes. What you then have is a 1 ohm resistor in series with a parallel combination of the 9 ohm resistor and whatever load you connect. You're ability to deliver power to the load will depend on the 1 ohm and 9 ohm resistor too much for it to be a versatile and reliable power delivery mechanism.

This type of problem doesn't really tell us anything except how voltage division works. There are not many practical applications of connecting these resistors to a 240V supply. The only reason I can think someone would purposely connect a resistor in series or parallel with a load is for motor braking or soft starting. I would be interested to see other practical applications.

 

[gar]

140416-1422 EDT

drktmplr12:

A practical application is an oscilloscope X10 probe. Nine megohms in the probe working with a 1 megohm input resistance at the scope. Also in combination with the resistive divider is a capacitive divider to provide good frequency response.

Or a probe I made with 50,000 V input capability to measure the applied voltage to a spark plug. This used a 500 megohm resistor. Frequency response into the kHz range was required.

.

 

[drktmplr12]

gar:

Interesting, I did not expect someone to entertain my request. Thanks for that!

In school, we focused mainly on applications of integrated circuits. I've been working in industrial power distribution systems for water treatment plants for a while now, so I don't think of these things! I failed to consider that integrated circuits and interfaces to them are made of discrete components and voltage/current dividers are rampant at that level of design.

Sounds like a very interesting problem with measuring the voltage applied to a spark plug.

Wonder if OP will return? :?

edit: grammar

 

[renosteinke]

Sure you can ... but ... you won't get the values you expect!

What you describe is literally the very first problem given in the very first class aspiring engineers take in engineering school. The joker in the deck is that AC is always changing, so a proper solution to the problem required the use of calculus. Old-fashioned DX/DY stuff.

Oh .. and btw ... you're liable to accidentally create a 'linear amp,' radiating all manner of electronic noise and pissing off everyone with a radio within blocks.