Question

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)?

Answer 1

r = 5.5 ohms

Step-by-step 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