Real Circuit, Thevenin problem!

Thread Starter

treyland123

Joined Oct 26, 2016
1
Hi,

I need a little help with understanding this problem and how it should be laid out (attached below).

In the question, it says the source impedance is 50 ohm and voltmeter impedance is 1M ohm.
Does this simply mean that I can model the system with a 50 ohm resistor in series after the source, and 1M ohm in series with the voltmeter?
And thus solve with Thevenin equivalence?

Thank you!

1.jpg
 

WBahn

Joined Mar 31, 2012
29,979
Hi,

I need a little help with understanding this problem and how it should be laid out (attached below).

In the question, it says the source impedance is 50 ohm and voltmeter impedance is 1M ohm.
Does this simply mean that I can model the system with a 50 ohm resistor in series after the source, and 1M ohm in series with the voltmeter?
And thus solve with Thevenin equivalence?

Thank you!

View attachment 114329
An ideal voltmeter has an infinite resistance, so putting the 1 MΩ in series with it would change nothing. So how else might it be modeled? Think of the case where the voltmeter is directly across the source and both have the same resistance. You would want the voltmeter to read half the source voltage, so your model must reflect that.

As for whether you use Thevenin or some other analysis technique, that is entirely up to you (subject, of course, to whatever constraints are placed on how you are expected to solve circuits on this assignment).
 

DGElder

Joined Apr 3, 2016
351
Given that this represent a "real circuit": the elements in the circuit are specified to 1 signficant digit. The impedance of the voltmeter is 3 orders of magnitude greater than the output impedance of the circuit; so you can assume that the impedance of the voltmeter is infinite. Then you just need to model the ideal 10V source with a 50 ohm series resistor in series with resistors 1K||1K which forms a voltage divider to give the output voltage.
 

ci139

Joined Jul 11, 2016
1,898
i again would understand the "in the circuit" voltage the voltage* on terminals of the voltage source and thus also on the voltmeter´s -- whatever their impedances are the meter showing exactly that* or then some minor less due the impedance of the real wires assuming them not some sort of superconductors
 

WBahn

Joined Mar 31, 2012
29,979
Given that this represent a "real circuit": the elements in the circuit are specified to 1 signficant digit. The impedance of the voltmeter is 3 orders of magnitude greater than the output impedance of the circuit; so you can assume that the impedance of the voltmeter is infinite. Then you just need to model the ideal 10V source with a 50 ohm series resistor in series with resistors 1K||1K which forms a voltage divider to give the output voltage.
The "real" only indicates that the source and the meter are not ideal.
 
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