Fun little brain teaser

Thread Starter

KL7AJ

Joined Nov 4, 2008
2,229
I ran across this ages ago, and I present it to each new electronics class as an exercise.

You have a black box with four leads coming out of it. You apply a voltage (ac or dc) to any two adjacent leads. The two remaining leads output precisely 1/4 of the applied voltage. What is in the box?


Eric
 

strantor

Joined Oct 3, 2010
6,798
a wheatstone bridge with an inverted side? Like this:

the 12V source represents your applied voltage, and points C & D represent your output leads. Potential difference between C & D is 4V (1/3 applied voltage in this case).
 

KJ6EAD

Joined Apr 30, 2011
1,581
I agree in principal but don't the resistors have to hold the ratio X:3X to satisfy the 1/4 voltage requirement?
 

strantor

Joined Oct 3, 2010
6,798
I agree in principal but don't the resistors have to hold the ratio X:3X to satisfy the 1/4 voltage requirement?
The values can be changed to make it hold a 1/4 voltage when voltage is applied from opposite corners, but when it is applied on the same side, it no longer works. So I proved myself wrong. I'll think about more tomorrow. Good night.
wheatstone.png
 

AnalogKid

Joined Aug 1, 2013
11,056
You have a black box with four leads coming out of it. You apply a voltage (ac or dc) to any two adjacent leads. The two remaining leads output precisely 1/4 of the applied voltage. What is in the box?
If the four leads are in a row, does the outside pair constitute an adjacent pair as if the four leads were in a box?
Or does four-in-a-row intentionally have one fewer possible connection?

ak
 
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