# Thevenin and Norton Equivalent Circuits

## Thevenin Ideal Voltage Source

$$ R _ {out} = \frac{V_{th}}{I_{SC}} = \frac{V_{OC}}{I_{SC}} $$

Where the open circuit voltage, \( V_{OC} = V _{th} \), and \( I_{SC} \) is the short circuit current.

## Norton Ideal Current Source

$$ R _ {out} = \frac{V_{OC}}{I_{SC}} = \frac{V_{OC}}{I_{N}} $$

Where the short circuit current, \( I_{SC} = I _N \), and \( V_{OC} \) is the open circuit voltage.

eg. A practical voltage source provides \( 1.5V \) on open-circuit. If an ammeter with an effective internal resistance of \( 5 \Omega \) is connected between its terminals, a current of \( 14 mA \) is registered. Construct an equivalent circuit for the voltage source.

The equivalent Thevenin circuit consists of an ideal voltage source connected in series with an internal resistance. Since the open-circuit voltage is equal to the ideal source voltage, the internal resistance can be calculated by applying Ohm's law to the circuit when the ammeter is connected.

$$ V= IR $$

$$ 1.5 = 14 \times 10 ^{-3} (R_{int} +5) $$

$$ R_{int} = \frac{1.5}{14 \times 10 ^{-3}} -5 = 102.1 \Omega $$

The equivalent circuit is then as follows: