Need more help with circuits homework

[JohnCU]

Is the interconnection of ideal sources in the circuit in Fig. P2.5 valid? Explain.

figure 2.5

I say no, because the bottom section doesn't have the same voltage at each terminal. Now, if the polarity of the 18V source was switched, then I think it might work...I dont know I'm so confused in this class. All the positive and negative sign notations and everything else is clogging up my mind.

 

[TuxDave]

Assume the bottom of the circuit is ground or 0V. Above the 18V supply will be -18V, above the 7V supply will be +7V. No problem so far!

When looking for conflicting ideal sources, you need to look for two things. Will there be a situation where one node has to be 2 voltages. Like if you had a 1V and a 2V supply hooked up in parallel. The second is if there is a branch that has two different currents running through it like a 1A current source feeding directly into another 1A current source facing the opposite direction.

If you need more help, PM me or reply here.

 

[JohnCU]

Oh, so the bottom of the circuit doesn't have to be 18V or -7V...

 

[TuxDave]

Originally posted by: JohnCU
Oh, so the bottom of the circuit doesn't have to be 18V or -7V...

Nope, they just give you the difference such that from the positive end of the voltage source to the negative end of the voltage source, there will be a difference of X volts.

 

[JohnCU]

Okay, there are 2 more questions associated with this figure.

b. Identify which sources are developing power and which are absorbing.
I think that since the current is going through a drop in voltage over the 7v source that it's positive which means power is being absorbed, and since it is going through a rise over the 18v source that it's negative so that means power is being delivered...?

and c. Verify that the toal power developed in the circuit equals the total power absorbed.
I'm not sure how to verify this...18v*5ma != 7v*5mA

 

[Triumph]

So that means there is an 11 volt difference across the current source?

 

[JohnCU]

Originally posted by: Triumph
So that means there is an 11 volt difference across the current source?

Wait, doesn't it mean there is a -25 volt difference? -18 - 7 = -25

 

[TuxDave]

Originally posted by: JohnCU
Okay, there are 2 more questions associated with this figure.

b. Identify which sources are developing power and which are absorbing.
I think that since the current is going through a drop in voltage over the 7v source that it's positive which means power is being absorbed, and since it is going through a rise over the 18v source that it's negative so that means power is being delivered...?

and c. Verify that the toal power developed in the circuit equals the total power absorbed.
I'm not sure how to verify this...18v*5ma != 7v*5mA

b. Easy way to check for delivery or absorption. Find the way the current is flowing through either a voltage source or current source. If there is a voltage drop, then that device is absorbing. If there is a voltage boost, the device is delivering. You're right on the 7v part but wrong on the 18V one. Look at the positive and negative terminals. You'll see that through the 18V (since current is going UP through it), there is a DROP of 18V meaning it's absorbing energy.
c. You forgot the current source. Use the same methodology as above. The current is flowing from left to right, so look at the voltages on either side. Does the voltage DROP or GO UP? Then you can determine whether or not it's absorbing or delivering.

Originally posted by: JohnCU

Originally posted by: Triumph
So that means there is an 11 volt difference across the current source?

Wait, doesn't it mean there is a -25 volt difference? -18 - 7 = -25

Yes

 

[TuxDave]

.....

 

[MichaelD]

Wow. Just wow. I know what a zeneer diode does...can I play too?

 

[dowxp]

not sure if this is right since i hate circuits.

http://downloadanime.org/misc/circuit.pdf

 

[TuxDave]

Originally posted by: dowxp
not sure if this is right since i hate circuits.

http://downloadanime.org/misc/circuit.pdf

Using KVL with an ideal current source in the loop is kinda pointless since it has no intrinsic voltage across it. So you can always assign whatever voltage across it to make the loop back to 0V. Other than that, the power part is right.

Oh.... nice tablet PC.... (if I'm guessing correctly)