# 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: 