During the engineering course, I had read and seen the Thevenin equivalent circuit theory and example of the circuit, I completely forgot it. As I now read more often circuit and circuit schematic to find out how they work, I have started to understand a bit more and can figure out why and which circuit part or part is being used and why. The Thevenin equivalent circuit can be used wherever you see a source connected across a voltage divider circuit. And a voltage divider circuit is common in electronics circuit of any application. So using Thevenin equivalent circuit we can reduce the circuit which helps to understand the circuit better.
The diagram below shows how one can replace a voltage divider source with thevenin equivalent circuit.
Here is why,
The Voltage Divider Circuit: Lets understand this first.
The output voltage \(V_{out}\) is the voltage across \(R_2\),
\(V_{out} = IR_2\) ------>(1)
The current is,
\(I = \frac{V_{in}}{R_1+R_2}\)
using this in (1),
\(V_{out} = \frac{R_2}{R_1+R_2} V_{in}\)
Ok, what about Thevenin Theorem? The Thevenin Theorem states that, "Any linear electrical network containing only voltage sources, current sources and resistances can be replaced at terminals A–B by an equivalent combination of a voltage source Vth in a series connection with a resistance Rth."
How to verify mathematically for the voltage divider circuit? Otherwise how to find the Thevenin equivalent circuit \(V_{th}\) and \(R_{th}\).
The \(V_{th}\) is the open circuit voltage,
\(V_{th} = V(open circuit) \) ------->(1)
The \(R_{th}\) is the ratio of open circuit voltage V(open circuit) and short circuit current I(short circuit). That is,
\( R_{th} = \frac{V(open circuit)}{I(short circuit)} \) ----->(2)
The open circuit voltage V is,
\(V = \frac{R_2}{R_1+R_2} V_{in} \)
and the short circuit current I is,
\(I = \frac{V_{in}}{R_2} \)
And so,
\(V_{th} = \frac{R_2}{R_1+R_2} V_{in} \)
and,
\( R_{th} = \frac{R_1R_2}{R_1+R_2} \)
