Identifying a Voltage Divider Circuit: Tips for Quick Recognition

[kusiobache]

How does one tell if a circuit is a voltage divider or not, without working out the circuit? Because in class people just look at the circuit and say "well that's a voltage divider so..." and then procede to solve the problem.

So, how does one tell if a circuit is a voltage divider just by looking at it?

 

[Redbelly98]

A voltage divider consists of 2 resistors in series, with an input voltage applied across both of them. If the output voltage is taken at the point in between the two resistors, then it is a voltage divider.

 

[kusiobache]

That makes a lot of sense. Thank you.

It doesn't need to be two resistors though, does it? It can be a resistor and inductor, or resistor, can't it? And what if a resistor or inductor (pure) came first. They don't have any real resistance (it's imaginary), and don't dissipate power, so would it still be a voltage divider?

 

[davenn]

That makes a lot of sense. Thank you.

It doesn't need to be two resistors though, does it?

well yes else you are not going to get a division of the voltage
will qualify that by saying ... anything that has resistance. eg there will be a very small resistance and hence very small voltage drop across
say the collector and emitter junction of a transistor

It can be a resistor and inductor, or resistor, can't it?

unless the inductor has a very large number of turns so as to give it some significant resistance then there isn't going to be much of a voltage drop across it

And what if a resistor or inductor (pure) came first.

a simple circuit ... +V line to inductor to resistor to 0V line
pick on an average inductor of a few turns or so resistance of the coil somewhere
between 0 and 1 ohm. If it was first in the cir4cuit then the voltage appearing on the side connected to the resistor would be practically the same as the supply voltage and ALL the voltage drop would be across the resistor.
if the resistor was first and inductor between the resistor and 0V. Because the resistance of the inductor is basically 0 Ohms, then that oV point is then also on the resistor/inductor connection point. So once again the the complete voltage drop is across the resistor

They don't have any real resistance (it's imaginary), and don't dissipate power, so would it still be a voltage divider?

of course it has to be real! you can't get a voltage drop across something that doesn't exist
and yes it will dissapate power using W = I x R ie. the risestance of the resistor x the current through it

cheers
Dave

 

[kusiobache]

well yes else you are not going to get a division of the voltage
will qualify that by saying ... anything that has resistance. eg there will be a very small resistance and hence very small voltage drop across
say the collector and emitter junction of a transistor



unless the inductor has a very large number of turns so as to give it some significant resistance then there isn't going to be much of a voltage drop across it



a simple circuit ... +V line to inductor to resistor to 0V line
pick on an average inductor of a few turns or so resistance of the coil somewhere
between 0 and 1 ohm. If it was first in the cir4cuit then the voltage appearing on the side connected to the resistor would be practically the same as the supply voltage and ALL the voltage drop would be across the resistor.
if the resistor was first and inductor between the resistor and 0V. Because the resistance of the inductor is basically 0 Ohms, then that oV point is then also on the resistor/inductor connection point. So once again the the complete voltage drop is across the resistor



of course it has to be real! you can't get a voltage drop across something that doesn't exist
and yes it will dissapate power using W = I x R ie. the risestance of the resistor x the current through it

cheers
Dave

You're answers are very clear and straightforward, so thank you very much. I appreciate the help man

 

[davenn]

have a look at this pic and voltages of your 2 situations...
Thats assuming as in the comments in the previous post that the resistance of the inductor is near zero
as to not have a significant voltage drop for the sake of this discussion

cheers
Dave

 

[vk6kro]

Since you have raised the question of inductors, you might like to see what happens if an AC waveform is applied to the circuit instead of a battery.

Suppose a 100 volt peak signal at 1000 Hz is applied to the series circuit of a 0.1 uH inductor and a 1000 ohm resistor.
The resistor would have 84.67 volts peak across it and the inductor would have 53.2 volts peak across it.
If you add these up, you get 137.87 volts. But we only started with 100 volts.

However, 84.67 squared plus 53.2 squared gives 10178 and the square root of this is 100 volts. Just what we started with.

You can put the voltages on a right angled triangle and work out unknown voltages from Pythagoras's Theorem.
You can also put the resistor and the reactance of the inductor on a right angled triangle and get the impedance of the circuit on the long side opposite the right angle.

So, this still works as a voltage divider, but the calculation is not as simple as with batteries.

Inductance and reactance and capacitance are all very real and far from imaginary.

 

[Redbelly98]

It doesn't need to be two resistors though, does it? It can be a resistor and inductor, or resistor, can't it? And what if a resistor or inductor (pure) came first. They don't have any real resistance (it's imaginary), and don't dissipate power, so would it still be a voltage divider?

Since you have raised the question of inductors, you might like to see what happens if an AC waveform is applied to the circuit instead of a battery.

vk6kro raises a good point about AC waveforms, and I oversimplified matters when I said a divider consists of 2 resistors in series. Instead of a resistor, it could be any combination of components that has a complex impedance (i.e. linear devices: resistors, inductors, and/or capacitors), in series with another combination of such components. As long as you can calculate impedances, the voltage divider concept is applicable.