quick question on an electronic paper the output of a circuit is measured with a digital multimeter, before and after a load resistance. the meter input resistance is about 10^8. the meter reading drops to 10% of its original value when a load of 5kÙ is added. discuss how the measurement will be affected if the load was 1)10Ù 2)500kÙ anyone help please...
The impedance (resistance) of the source is X ohms, so with the 5k load 1/10 of the source voltage is dropped across the 5k, the other 9/10 across X. You can do the rest now.
so basically if it were 500kohm it would be 0.5% drop across that and the rest accross the volt meter? i am a bit of a dummy on this front...
Don't feel bad, all Doncaster people are like that... You can forget about the meter resistance, at 10^8 ohms it doesn't affect things to any extent. I make the source resistance 9*5k=45k, so 45k in series with the 500k load will leave 500/545 *100% for the meter to read. 45k in series with a 10R load will only leave 10/45010*100% registering on the meter.
is it that the higher the load resistance of the circuit, the more significant the internal resistance of the multimeter will become in contributing to the measurement error of the circuit output voltage. is there a forumla or summat? where r u in s. yorks btw
It can be a problem if there's another highish resistance in the circuit apart from the resistance the multimeter is connected across, like the example you've got. You can work out what happens with Ohms Law. Code: 12v----Rs------Rl-------0v |--Rm--| The load resistance with a meter across it (Rc) becomes a different value, 1/Rc = 1/Rl + 1/Rm where Rl is load resistance and Rm is meter resistance. So Rs and the combined Rl || Rm (=Rc) split the 12v differently to just Rs and Rl. Again Ohms law will show you what happens. No meter, voltage across Rl = 12 x Rl/(Rl + Rs) Metered, voltage across Rl= 12 x Rc/(Rc + Rs) With a modern digital meter, Rm is so high it makes very little difference (10^8 = 100,000k), unless Rl is also extremely high. But with cheap needle voltmeters Rm is only a few thousand ohms so gives big errors on high resistance circuits. --- cpemma@mexbro.couk
As the question says, The output circuit of a pre-amp, cd-player, soundcard or lots of other circuits may have a fairly high source resistance. So you have to use a higher load resistance to get plenty of signal voltage from the source to the load.
