When a circuit is connected a load, part of the voltage signal will be lost on i

Question

When a circuit is connected a load, part of the voltage signal will be lost on its own output resistance Ro. Find the voltage loss (in percentage) of the circuit below, when the signal Vs is transmitted to the load R_1, given the following input/output impedance: Ro = 50 ohm; R_l = 1 M ohm; Ro = 50 Q; R_l = 1 K ohm; Ro = 1 K ohm; R_l = 10 K ohm. What conclusion can you draw about input/output impedances from this question? As we've discussed in class, the thermistor is an extremely sensitive device because its resistance changes rapidly with temperature. We've also discussed the problems and disadvantages of such. This non-linearity also affects the sensitivity of the thermistor as a function

Explanation / Answer

Basically, Ro is the internal resistance of the source and the voltage drop across the internal resistance is always considered as a loss. The calculation in detail is shown below:

If i(t) is the current flowing in the circuit, then i(t)=Vs/(Ro+RL)----->(1)

(a) Ro=50 Ohms and RL=1 Mega Ohms. Then voltage developed across the internal resistance V=i(t)XRo=Vs.Ro/(Ro+RL) and this voltage V is considered to be a loss. Putting the value of Ro and RL, the percentage loss becomes: (V/Vs)X100%=100X Ro/(Ro+RL) =0.005%

(b) Ro=50 Ohms and RL=1 Kilo Ohms. Then voltage developed across the internal resistance V=i(t)XRo=Vs.Ro/(Ro+RL) and this voltage V is considered to be a loss. Putting the value of Ro and RL, the percentage loss becomes: (V/Vs)X100%=100X Ro/(Ro+RL) =4.76%

(c) Ro=50 Ohms and RL=10 Kilo Ohms. Then voltage developed across the internal resistance V=i(t)XRo=Vs.Ro/(Ro+RL) and this voltage V is considered to be a loss. Putting the value of Ro and RL, the percentage loss becomes: (V/Vs)X100%=100X Ro/(Ro+RL) =0.476%

From these three results, it can be concluded that the higher the value of the load RL compared to the internal resistance of the souce Ro, lesser is the percentage of voltage loss across the source resistance.

The matlab code is given below:

Vs=input('Enter the value of the source voltage') %take any value of input, keep it fixed
Ro=input('Enter the value of internal resistance of source')
RL=input('Enter the value of load resistance')
i=Vs/(Ro+RL);
V=i*Ro;
error=(V/Vs)*100;
disp('%loss in voltage is')

disp(error)

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