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Five Resistors

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Five Resistors

Postby MJG » Fri Apr 17, 2015 6:47 am

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Can anyone tell me the voltages on either end of R3 (the horizontal one in the middle)?

I would greatly like to know how you arrived at the answer.

TIA
MJG
 
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Re: Five Resistors

Postby pebe » Thu Apr 23, 2015 4:32 pm

I make the answer 8.57V across the resistor.

Using Thevenin, R1 and R4 in the circuit can be replaced by the equivalent of the two resistors in parallel, and fed from the open circuit voltage at their junction. So we have 3.2ohms fed from 20 x 4 / 20 = 4V. Similarly, R2 and R3 can also be replaced by a single resistor of 3.2ohms and fed from 20 x 16 / 20 = 16V.

So in effect we have R3 (16ohms) in series with 2 x 3.2V with 12V across them. So the voltage across R3 = 12 x 16 / 22.4 = 8.57V, ie. 14.285 V at one end and 5.715V at the other.
pebe
 
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Re: Five Resistors

Postby MJG » Fri Apr 24, 2015 5:07 am

Thanks so much. After poking around with Thevinizing, I finally got it!

The circuit is actually a "Bridged T" attenuator where R3 is a 16ohm speaker and the other four are power resistors. Applying the full 20V into the lone 16ohm speaker yields 25 watts, while using this attenuator circuit reduces the output to 12.75W which is about half power (but not sound level), non?

Thanks again...
MJG
 
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Joined: Fri Apr 17, 2015 6:39 am


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