# Homework Help: Connecting a load to a voltage amplifier

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1. Jul 4, 2016

### bl965

• Member advised to use the homework template for posts in the homework sections of PF.
Hi. This is a problem from an electronics textbook:
The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 k is connected. What is the value of the amplifier output resistance?

How is there an output voltage without there being a load? 80% compared to what?
Thank you.

2. Jul 4, 2016

### phinds

Why should there NOT be an output voltage with no load? Do you think, for example, that a power supply has an output of zero volts until you hook it up to something?

By the way, the implication of your question as stated is not that the output voltage is zero but that there isn't one at all, which is impossible. I assume you intended to imply the question "why is there an output voltage of zero volts with no load?"

3. Jul 4, 2016

### Merlin3189

If you think of the amplifier as a box with two terminals labelled "output", then with no load attached, you could measure the voltage across the output terminals.
Then you could attach a 1k resistance across the terminals and measure the voltage again.
The question assumes that the voltage is stable or fixed (probably AC with constant peak to peak voltage), otherwise you could not meaningfully compare voltages at different times. You could even imagine it was a fixed DC voltage, like a battery. The result here would be the same. It does not matter what that voltage is - if it was 10V with no load it would be 8V with the 1k load, if it were 5V with no load it would be 4V with the 1k. You could assume an arbitrary voltage or just call it U.

4. Jul 4, 2016

### bl965

So 10 Volts with no load would only have the internal resistance in the circuit, then would the drop across the internal resistance be 10 V? If there was a load output would be measured across the load resistance, but where is output measured without a load?

5. Jul 4, 2016

### Merlin3189

Try drawing a circuit maybe?
If you have the box with no load, then there is no current flow, so no voltage drop in the internal resistance. !0V at the output means 10V source inside.
Then when you connect a resistance, some current flows and there is now some voltage across the internal resistance.

Last edited: Jul 4, 2016
6. Jul 4, 2016

### bl965

Got it!
U 1000/(1000+R) = 0.8U
R = 250
Thank you!