Basic Electronic Question

[elster]

I think I am being a bit dumb here.



If the Voltmeter has a resistance on 1800 Ohms and the battery has an internal resistance of 50 Ohms are they treated as all 3 in parallel.

Or is the internal resistance of the battery treated as series to the voltmeter and resistor?

[twister]

Redraw the circuit with the internal resistances shown as discrete elements and it should be clear - the voltmeter internal resistance parallels the load resistance, and the battery internal resistance is in series with this parallel combination.

[Oldandslow]

^What he said.

[wackojacko]

twister said:

Redraw the circuit with the internal resistances shown as discrete elements and it should be clear - the voltmeter internal resistance parallels the load resistance, and the battery internal resistance is in series with this parallel combination.

Beat me to it Exactly as said above ^

[r1ot]

Not sure it matters, I'd be seeing 12V on the voltmeter and the resistor would be dissipating a healthy 720mW....

[wackojacko]

You would see 12v on the volt meter as it is in the drawing but effectively it would help find the voltage drop across the resistor. V= I x R the voltage drop should equal 12v in a series circuit where as parallel will be some return so the drop will never be equal to the supply.

The answer to your question the drawing is showing no out put i.e a motor or lamp, i take it your using the voltmeter as the generic output ? if so it would be a parallel circuit with but the resistor would need to be on the positive feed side of the out put for it to be of much use to the out put. or it could be used to dampen the back current. hope this helps mate.

Edited by wackojacko on Saturday 30th January 23:45

[Ozzie Osmond]

r1ot said:

Not sure it matters, I'd be seeing 12V on the voltmeter and the resistor would be dissipating a healthy 720mW....

I'm with you on that one.

Maybe we haven't understood the purpose of the question?

[Pesty]

not sure I understand what he is asking either

OT
I get I = 0.06A

after all these years my o-level in physics pays off woo woo. I used an equation to work somthing out.

I'm feeling proud of myself if I am wrong keep it to yourself

[elster]

Well no as you haven't taken into account the internal resistance of the battery.

I have done it with the battery in series and the answer is 9.39V across the resistor.

If they were all in parallel then the answer would be 12V.

my workings here:

Resistors in parallel
1/R =1/200 + 1/1800 so R=180 Ohms

Total R = 180 + 50 = 230

V/R = I = 12/230 = 0.052A

V1 across 50 ohm resistor of battery = 0.052 x 50 = 2.61V
V2 across 180 Ohm resistors in series = 0.052 x 180 = 9.39V

As the potential difference of them in parallel is the same across both resistorws the answer would be 9.39V.

[Dave_L]

I think perhaps the mistake you are making here is the actual internal resistance of the battery. For a typical lead acid battery (I'll assume that since it's 12V ), the internal resistance is of the order of 50 milli-ohms not 50 ohms. Thus it makes practically bugger all difference in the real world.

[spikeyhead]

elster said:

Well no as you haven't taken into account the internal resistance of the battery.

I have done it with the battery in series and the answer is 9.39V across the resistor.

If they were all in parallel then the answer would be 12V.

my workings here:

Resistors in parallel
1/R =1/200 + 1/1800 so R=180 Ohms

Total R = 180 + 50 = 230

V/R = I = 12/230 = 0.052A

V1 across 50 ohm resistor of battery = 0.052 x 50 = 2.61V
V2 across 180 Ohm resistors in series = 0.052 x 180 = 9.39V

As the potential difference of them in parallel is the same across both resistorws the answer would be 9.39V.

This is right.

[elster]

spikeyhead said:

elster said:

Well no as you haven't taken into account the internal resistance of the battery.

I have done it with the battery in series and the answer is 9.39V across the resistor.

If they were all in parallel then the answer would be 12V.

my workings here:

Resistors in parallel
1/R =1/200 + 1/1800 so R=180 Ohms

Total R = 180 + 50 = 230

V/R = I = 12/230 = 0.052A

V1 across 50 ohm resistor of battery = 0.052 x 50 = 2.61V
V2 across 180 Ohm resistors in series = 0.052 x 180 = 9.39V

As the potential difference of them in parallel is the same across both resistorws the answer would be 9.39V.

This is right.

Cheers. I was wondering if it was a trick question and they were all really in parallel.

[10 Pence Short]

Why don't you just send your XBOX back to Microsoft like normal people?

Edited by 10 Pence Short on Sunday 31st January 11:26

[twister]

Dave_L said:

I think perhaps the mistake you are making here is the actual internal resistance of the battery. For a typical lead acid battery (I'll assume that since it's 12V ), the internal resistance is of the order of 50 milli-ohms not 50 ohms. Thus it makes practically bugger all difference in the real world.

Considering that in the real world a voltmeter with an internal resistance of 1K8 would be found only in the pile of stuff labelled "fix it or dump it, but don't use it", I suspect the problem is using contrived values rather than realistic ones in order to better highlight the effects of internal resistances...

[elster]

twister said:

Dave_L said:

I think perhaps the mistake you are making here is the actual internal resistance of the battery. For a typical lead acid battery (I'll assume that since it's 12V ), the internal resistance is of the order of 50 milli-ohms not 50 ohms. Thus it makes practically bugger all difference in the real world.

Considering that in the real world a voltmeter with an internal resistance of 1K8 would be found only in the pile of stuff labelled "fix it or dump it, but don't use it", I suspect the problem is using contrived values rather than realistic ones in order to better highlight the effects of internal resistances...

Exactamundo.

[spikeyhead]

elster said:

twister said:

Dave_L said:

I think perhaps the mistake you are making here is the actual internal resistance of the battery. For a typical lead acid battery (I'll assume that since it's 12V ), the internal resistance is of the order of 50 milli-ohms not 50 ohms. Thus it makes practically bugger all difference in the real world.

Considering that in the real world a voltmeter with an internal resistance of 1K8 would be found only in the pile of stuff labelled "fix it or dump it, but don't use it", I suspect the problem is using contrived values rather than realistic ones in order to better highlight the effects of internal resistances...

Exactamundo.

It's a simple question to show some principlpe.

I'm currently playing with devices with DC input impedances in Giga-ohms, and others with input and output impedances that are a small fraction of an ohm. If you don't understand the basic principal that measurements affect the circuit being measured then the results would be garbage.

[Ozzie Osmond]

spikeyhead said:

If you don't understand the basic principal that measurements affect the circuit being measured then the results would be garbage.

Yeah, but it's a bit of a nonsense question because in the real world for practical purposes,

a) the resistance of the meter is infinite, and
b) the resistance of the battery is zero.

Otherwise nobody would ever be able to measure anything. For instance, how could you ascertain the correct internal resistance of either the meter or the battery?

[spikeyhead]

Ozzie Osmond said:

spikeyhead said:

If you don't understand the basic principal that measurements affect the circuit being measured then the results would be garbage.

Yeah, but it's a bit of a nonsense question because in the real world for practical purposes,

a) the resistance of the meter is infinite, and
b) the resistance of the battery is zero.

Otherwise nobody would ever be able to measure anything. For instance, how could you ascertain the correct internal resistance of either the meter or the battery?

Did you not read the rest of my post. Try designing a mobile phone, where towards the end of the battery life it has a significant internal resistance. Ignore that resistance and you won't have a device that customers will buy.

The numbers in the example used aren't realistic in a modern situation, however there's many applications where it's very difficult to measure things accurately. Try measuring the input impedance of a FET input op-amp

http://en.wikipedia.org/wiki/Operational_amplifier says

Finite input impedance
The input impedance of the operational amplifier is defined as the impedance between its two inputs. It is not the impedance from each input to ground. In the typical high-gain negative-feedback applications, the feedback ensures that the two inputs sit at the same voltage, and so the impedance between them is made artificially very high. Hence, this parameter is rarely an important design parameter. Because MOSFET-input operational amplifiers often have protection circuits that effectively short circuit any input differences greater than a small threshold, the input impedance can appear to be very low in some tests. However, as long as these operational amplifiers are used in a typical high-gain negative feedback application, these protection circuits will be inactive and the negative feedback will render the input impedance to be practically infinite. The input bias and leakage currents described below are a more important design parameter for typical operational amplifier applications.

So you're trying to measure something that's nearly infinite, but as real life doesn't do infinity, it's not infinite. How is that measured? Without some basic grounding in measurement theory you won't get an accurate answer.