Designing a Cost-Effective Variable Frequency Generator for Microwave Cooking

[mearvk]

Suppose R and r in the above diagram are equal. Can you see that the voltages across them will also be equal because they both have the same current flowing in them?

Now the output of the power source is at the terminals marked A and B, so the voltage there has to be the internal voltage minus the voltage drop across the internal resistor, r.

So, if the resistors are equal, then the voltage across the resistor R will be half the internal voltage so you know what the size of the internal resistor is.

And that is what you were trying to measure.

I don't see how a resistor 'drops voltage'. A resistor drops current by increasing the resistance; the voltage remains constant.

From what I understand the voltage, minus any voltage regulators, is constant in this circuit.

 

[vk6kro]

You might have the wrong idea here.

Suppose you had 10 resistors in series across a 10 volt battery. ("In series" means they are connected end to end, so the same current goes through all of them.)

Let all the resistors be 470 ohms.

So, the total resistance is 4700 ohms. (when resistors are in series, their resistances add up like that).

The current will be ^E / _R so the current will be 10 / 4700 or 0.0021 amps.

Now, this flows in all the resistors so the voltage across each resistor is I * R or 0.0021 * 470 or 1 volt.

So, you can see that the resistors share the voltage according to their resistance.

Since you are obviously interested in Electrical things, could you get into a class and have someone teach you this stuff properly? Some of the concepts look simple, but they are full of traps and you could easily get the wrong ideas if you try to teach it to yourself.

 

[mearvk]

I don't understand the 'voltage across each resistor' as a concept.

Are you saying that if I test for voltage after the 10th resistor I will get 1v or 10v?

If I test for voltage after the first resistor do I get 9v or 10v?

Looking at I = V / R we'd expect that adding resistors to a 10v circuit would lower the current. It is not intuitive why the voltage would be shared or divided or anything else. The equation simply states that, assuming V is constant, that increasing R decreases I. Says nothing about V changing across changes in resistance and in fact says otherwise.

Intuitively:

After R1 you'd have I = V / R or 10v / 470 ohm or 0.02 amps.

After R2 you'd have I = V / R or 10v / 940 ohm or 0.01 amps.

But what you seem to be saying is that the equations should look like this:

After R1 you have 9v / 470 ohm worth of current.

After R2 you'd have 8v / 940 ohm worth of current.

This seems massively flawed since we'd have to constantly add voltage back into the circuits so that every time we encountered a resistor we didn't change the voltage from 3.3v or 5v to something unexpected.

 

[mearvk]

I think I see what you're saying now.

The more resistance, relative, to the sum of total resistance, the greater the voltage across that particular resistor. But what does this mean?

I struggle with the meaning of 'splitting' a voltage across several resistors. Can you explain that a bit?

 

[vk6kro]

If you measured the voltage with a voltmeter, there would be 1 volt across each resistor.

The voltages across all the resistors adds up to the supply voltage.

They don't have to all be the same resistance, so the voltage will divide itself in proportion to the resistors present.

If you measured the voltage across, say, 3 of the resistors then there would be 3 volts showing on the voltmeter. If you measured across all of them the voltage would be 10 volts.

The current in all the resistors is the same if they are in series.

 

[mearvk]

So I could take a 9v battery and via a small circuit, connect equal 9 resistors and splice into the portion between R5 and R6 to create a 5v supply?

 

[vk6kro]

Yes. You could do that.

You would take two wires out. As long as there were 5 resistors in between the two wires, there would be 5 volts out between the two wires.

It is usual to have one of the wires connected to the negative side of the supply..

 

[mearvk]

Cool.