# Voltage in a Resistor Square

Discussion in 'Homework Help' started by yan500, Jul 23, 2011.

1. ### yan500 Thread Starter Member

Jul 12, 2011
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0
Hey guys, I'm confused on question 24 on this worksheet: http://www.allaboutcircuits.com/worksheets/dc_sp.html

I keep getting 6.77V as the answer, but it's wrong. My process for solving this problem is setting it up as a circuit schematic and then just calculating how much voltage goes through the 3k3 resistor (3.89V) and then I get stuck.

How should I tackle this problem?

2. ### praondevou AAC Fanatic!

Jul 9, 2011
2,936
488
Forget the 3k3 resistor.

looking from the power supply, the 4k7 resistor is in series with the 1k5 and the 2k7 resistor.

4k7 +1k5 gives you 6k2.

That's 6k2 in series with 2k7. (=8k9)

That's 10.5V over 8k9 and you want to know the voltage drop over 6k2.
Applying rule of three gives you the right answer. 6k2 x 10.5V / 8k9 = 7.31V

yan500 likes this.
3. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
the 3k3 resistor is a "red herring".
Since the power supply output is 10.5v, the voltage from point A to point B is 10.5v.
The current through that 3.3k resistor does not matter for this problem. If you wish, calculate the current flow through it; I=E/R - but don't use the result for anything else in the problem.

Then, add up the other three resistors, and calculate what the current flow will be from point A to B.
Then multiply that current by the resistance between points A and D.

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4. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Couldn't have asked for a better explanation guys, thanks!

5. ### BillB3857 Senior Member

Feb 28, 2009
2,399
348
"the 3k3 resistor is a "red herring". I havent heart that term since AFUA school at Memphis in 1962!!!

The instructors there had at least two questions with red herrings on each weekly test. Thanks for the memories.