# Quiz help

Discussion in 'Homework Help' started by jaymo3141, Feb 17, 2014.

1. ### jaymo3141 Thread Starter New Member

Feb 16, 2014
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0
The steps I have done so far are:

1. Found Vx by ignoring the 4 Vx and doing a mesh on the loop which gives me 0.5 for Vx

2. Short circuit Voc

after this i'm stuck. Can someone solve this. My friend got -15v but i'm not sure if he's right. Either way i have to submit this quiz today so if -15v is wrong then i'm getting this question wrong. It would be helpful if someone walked me through the solution though. Thanks.

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2. ### tshuck Well-Known Member

Oct 18, 2012
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You can also solve it by realizing no current flows through the dependent source, so Vx is a voltage divider output. Then the dependent source makes its negative side 4*Vx lower than the voltage at Vx.

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Do you mean: Vx=4Vx+Voc
So Voc=Vx-4Vx
Is that what you mean?

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
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According to my math skills your Vx is wrong.

5. ### tshuck Well-Known Member

Oct 18, 2012
3,527
675
I am applying the definition of a voltage source: its positive side is X volts above its negative side, regardless of whether X is a constant or a function.

Since this doesn't load Vx (as there is no current through the dependent source), you can conclude that Voc = Vx - 4Vx because the voltage at the dependent source's positive side is the dependent variable. Since the dependent voltage source is 4Vx volts, its lower side will be 4Vx volts lower than the positive side. Same answer, but different approach.

Last edited: Feb 17, 2014
6. ### alfacliff Well-Known Member

Dec 13, 2013
2,449
429
5 volts. simple voltage devider. didnt even need a calulator.

7. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,975
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Is that your answer to Voc or Vx?

8. ### alfacliff Well-Known Member

Dec 13, 2013
2,449
429
voc. this durn thing wants a longer response.

9. ### shteii01 AAC Fanatic!

Feb 19, 2010
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I got -15 V for Voc.

10. ### studiot AAC Fanatic!

Nov 9, 2007
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The dependent voltage source is open circuit therefore passes no current in this circuit, as tshuck says.

So the 20k and 50k resistors form a simple potential divider, which gives Vx.

The voltage of the dependent source is stated to be 4Vx so its negative end is 20 volts below the node (+5) to which it is coupled.

I make Vx = +5 volts and Voc = -15 volts.

So now you have tshuck's mathematical method and a physical circuit explanation to compare.

11. ### WBahn Moderator

Mar 31, 2012
20,247
5,758
In other words, you couldn't solve it and so you just copied your friend's answer and submitted it as your own.

What's your school's policy on academic integrity?

12. ### jaymo3141 Thread Starter New Member

Feb 16, 2014
8
0
Thank you everyone. I made a stupid math error. Appreciate your help