Signal generator‘s internal impedance is purely resistive and has an open-circuit voltage of 3.5 V. When the generator is loaded with 22 ohms of resistance, its terminal voltage drops to 2.8 V. What is the generator’s output impedance (pure resistance)?

r = 5.5 ohms

Explanation:

Given:-

- The open circuit voltage, Vo = 3.5 V

- The terminal voltage, Vt = 2.8 V

- The load, R = 22 ohms

- The internal resistance = r

Find:-

What is the generator’s output impedance (pure resistance)?

Solution:-

- We see that the source Voltage (Vo) is not entirely used for the attached load. Some of the source voltage is dropped within due to the source internal resistance or impedance.

- The terminal voltage (Vt) is the amount of Voltage drop across the load. The current drawn by the load I can be determined by Ohm's Law:

                            Vt = I*R

                            I = Vt / R

                            I = 2.8 / 22

                            I = 0.12727 Amps

- Since, the attached load (R) and the pure impedance (r) of the source are in series. The current  ( I ) is constant across both. The potential drop across the pure resistance (r) can be determined from Ohm's law:

                          Vo - Vt = I*r

                          r = ( Vo - Vt ) / I

                          r = ( 3.5 - 2.8 ) / 0.12727

                          r = 5.5 ohms

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