# Students: help the Proffessor.

Discussion in 'Homework Help' started by MikeML, Dec 23, 2015.

1. ### MikeML Thread Starter AAC Fanatic!

Oct 2, 2009
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Ok students,

Here is a pop quiz:

Prob 1. You have a black-box with two terminals. The open circuit voltage is 20V. When a 50A load is connected across the two terminals, the voltage across the load is 14V. Draw the equivalent circuit of what is inside the box.

Prob 2. You have a black-box with two terminals. The open circuit voltage is 120V. When the output terminals are shorted, the current is 60A. When loaded with a variable resistor such that the output voltage is 14V, the load current is 53A. Draw the equivalent circuit of what is inside the box.

2. ### OBW0549 Well-Known Member

Mar 2, 2015
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The equivalent circuit is a 20V voltage source in series with a resistor of 0.12Ω. (Voltage drop under 50A load is (20-14) volts = 6 volts; 6 volts divided by 50 amps is 0.12Ω.)

Equivalent circuit is a 120V voltage source in series with 2 ohms resistance. This can be determined either by dividing the open circuit voltage by the short circuit current (120V/60A=2Ω) or by taking (120-14)V/53A = 106V/53A = 2Ω.

3. ### MikeML Thread Starter AAC Fanatic!

Oct 2, 2009
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How about a 60A current source shunted by 2Ω?

4. ### OBW0549 Well-Known Member

Mar 2, 2015
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Arrrrrgh!!!! I missed that one...

5. ### MikeML Thread Starter AAC Fanatic!

Oct 2, 2009
5,451
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I didn't mean to imply that your answer for case 2 is wrong. Do you agree that either answer is correct?

Back to case 1. With the information I presented, could that black box contain a 166.7A current source shunted with 0.12Ω?

6. ### WBahn Moderator

Mar 31, 2012
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4,920
Assumption: The circuit in the box is linear and can be adequately modeled by a Thevenin and/or Norton equivalent circuit.
Note that there are an infinite number of equivalent solutions.

Prob 1.

At 50 A the internal resistance of the Thevenin equivalent is dropping 6 V making Req = 6 V / 50 A = 120 mΩ.

The Norton equivalent circuit would therefore have a current source that is In = Voc / Re = 20 V / 120 mΩ = 166.7 A

Thus two equivalent solutions are a 20 V source in series with a 120 mΩ resistance or a 166.7 A source in parallel with a 120 mΩ resistance.

Prob 2.

This problem is over defined and therefore may not have a linear solution.

Using the first data point (Iss = 60 A):

At 60 A the entire 120 V is dropped across Req, therefore Req = 120 V / 60 A = 2 Ω.

The Norton equivalent circuit would be a 60 A source.

Using the second data point (Vout = 14 V, Iout = 53 A)

Req = (120 V - 14 V) / 53 A = 2 Ω

So the two data points are consistent.

Thus two equivalent solutions are a 120 V source in series with a 2 Ω resistance or a 60 A source in parallel with a 2 Ω resistance.

7. ### OBW0549 Well-Known Member

Mar 2, 2015
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Absolutely. The box could contain a 120V voltage source in series with 2Ω, or a 60A current source shunted by 2Ω, and you wouldn't be able to tell the difference from the outside.

Yup. It certainly could, and as above you wouldn't be able to tell the difference from outside the box.

8. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
You wouldn't be able to tell them apart based on voltage/current measurements. But if you monitored the heat coming off of the box you could distinguish between the two. Though you still couldn't determine what the actual circuit was, just rule some of the possibilities out.

9. ### Bordodynov Active Member

May 20, 2015
673
194
The voltage source with a resistor to an external load is equivalent to the current generator with a resistor. For external load is no difference. However, with no load current generator internal resistor dissipates more power. For a voltage source in such a case, however, power to the internal resistor is equal to zero.
It's like a parallel and serial stabilizers.

Last edited: Dec 24, 2015