IIT JEE , AIEEE Forum - Physics , Electricity

rock

sir,could u explain thevinin theorem with some applications. aseem aka rock

edison

In electrical circuit theory, Thevenin's theorem for electrical networks states that any combination of voltage sources, current sources and resistors with two terminals is electrically equivalent to a single voltage source V and a single series resistor R. For single frequency AC systems the theorem can also be applied to general impedances, not just resistors.

Any black box containing only voltage sources, current sources, and resistors can be converted to a Thévenin equivalent circuit.

Any black box containing only voltage sources, current sources, and resistors can be converted to a Thevenin equivalent

STEPS FOR SOLVING THE NETWORK USING THEVENIN'S THEOREM

Calculate the output voltage, V_AB, when in open circuit condition (no load resistor - meaning infinite resistance). This is V_Th.
Calculate the output current, I_AB, when those leads are short circuited (load resistance is 0). R_Th equals V_Th divided by this I_AB.

The equivalent circuit is a voltage source with voltage V_Th in series with a resistance R_Th.

Case 2 could also be thought of like this:

2a. Now replace voltage sources with short circuits and current sources with open circuits.

2b. Replace the load circuit with an imaginary ohm meter and measure the total resistance, R, "looking back" into the circuit. This is R_Th

Example of a Thévenin equivalent circuit

Step 0: The original circuit

Step 1: Calculating the equivalent output voltage

Step 2: Calculating the equivalent resistance

Step 3: The equivalent circuit

In the example, calculating equivalent voltage:

$V_\mathrm{AB} = {R_2 + R_3 \over (R_2 + R_3) + R_4} \cdot V_\mathrm{1}$

$= {1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \over (1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega) + 2\,\mathrm{k}\Omega} \cdot 15 \mathrm{V}$

$= {1 \over 2} \cdot 15 \mathrm{V} = 7.5 \mathrm{V}$

Calculating equivalent resistance:

$R_\mathrm{AB} = R_1 + \left ( \left ( R_2 + R_3 \right ) \| R_4 \right )$

$= 1\,\mathrm{k}\Omega + \left ( \left ( 1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \right ) \| 2\,\mathrm{k}\Omega \right )$

$= 1\,\mathrm{k}\Omega + \left({1 \over ( 1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega )} + {1 \over (2\,\mathrm{k}\Omega ) }\right)^{-1} = 2\,\mathrm{k}\Omega$

Lois

Can you please explain how is equivalent voltage (7.5 V) calculated here?

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