0
\$\begingroup\$

I have a hard time trying to understand why Thevenin’s Theorem is the way it is.

As per the diagram below, I understand why we take our “Thevenin resistance” that way. But as far as the voltage is concerned, I am completely lost. Why do we take the “Thevenin voltage” to be the voltage drop across the points A and B? By doing that, won’t the voltage drop across AB in the equivalent circuit be less than the original value, since part of the voltage must have been dropped across the “Thevenin resistance”?

This has been bothering me for two days now. Please help.

image

\$\endgroup\$
  • \$\begingroup\$ what is the voltage drop across a resistor in an open circuit? \$\endgroup\$ – jsotola Sep 29 '19 at 20:52
1
\$\begingroup\$

e= shows only the no-load voltage but not r.

With the voltage source Rs=0, r = R1//R3 +R2

\$\endgroup\$
0
\$\begingroup\$

Thevenin's theorem is used to model the circuit for different loading conditions. This allows you to find the current flowing in the circuit and the voltage drop in the load resistor when load is connected.

So Thevenin's voltage is the voltage across AB when no load is connected across AB. Connect a 1K ohm resistor at the output terminal. You can now see that the circuit is closed and the current flows. This current causes drop in the Rth so the output voltage is no longer Vth but less than Vth.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.