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A signal source that is most conveniently represented by its Th´evenin equivalent has vs = 10 mV and Rs = 1 k. If the source feeds a load resistance RL, find the voltage vo that appears across the load for RL = 100 k, 10 k, 1 k, and 100 . Also, find the lowest permissible value of RL for which the output voltage is at least 80% of the source voltage.

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Answer:

RL=100K → Vo=9.90 mV

RL=10K → Vo=9.09 mV

RL=1K → Vo=5 mV

RL=100 → Vo=909.09 μV

In order to obtain 80% of the power source we have to put a resistor of 4 KOhm.

Explanation:

Here we have a power source in serie with a resistor of 1K and  RL, in order to obtain the Vo voltage we have to apply the voltage divider rule, that states:

[tex]Vo=Vin*\frac{RL}{RL+R1} \\R1=1kOhm[/tex]

Substituing the resistor values of RL we obtained the following results:

RL=100K → Vo=9.90 mV

RL=10K → Vo=9.09 mV

RL=1K → Vo=5 mV

RL=100 → Vo=909.09 μV

In order to find the lowest value that gives us 80% of the source voltage we have to use the voltage divider rule again and make the Vo equal to 0.8 Vin:

[tex]0.8*Vin=Vin*\frac{RL}{RL+R1}[/tex]

The result of the last equation is 4000, so in order to obtain 80% of the power source we have to put a resistor of 4 KOhm.

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