Hello

i am solving dependent sources questions , i ran into a problem ...

please help , as in the image ...

[80]+[20+100+80]+[-260]+20 != 0

[80]+[20+100+80]+[-260]-20 = 0

don't we say that the current enters the [-] then [-2*[-10]] then 20W !!
help !! thanks a lot .

 

What is the value of the third resistor?

hgmjr

 

20 Ω .

you know man , v1 = 10 V , v2= 50V , v3 = 40V .

 

The two dependent sources are the power sources for the circuit = 280W total. The series resistors account for 200W loss and the other 80W is absorbed by the 40V source - the power sinks and sources do therefore balance.

 

The two dependent sources are the power sources for the circuit = 280W total.

don't sources sometimes deissipate power ? , how did you know that those two generate , according to the book methodology , it is 20 not -20 .


if the question asked for the two dependent sources power only , i will not bother sum to zero and i will be mistaken , what is that ? .

another guy somewhere said

Put a negative resistor in parallel with a constant voltage source and have some fun! Hey, just because the numbers are upside down doesn't mean it's incorrect -- it's quite possible to construct a negative resistance. If you can't imagine that, just wait 'til AC analysis literally turns your numbers sideways...

Tim

i don't understand him , what it has to do with this circuit .

 

Re-check your final dependent source polarities - the 2 bottom sources equivalents shown on your post will have the same polarities - i.e right side positive - not opposite as you have shown. This is based on the assumption that the initial source polarities on the starting circuit are as intended.

Ideal sources can only source or sink (absorb) power - not dissipate. There needs to be a resistive component (in the non-ideal source) for dissipation (losses) to occur.

 

the book gave the polarities above as drawn .

Ideal sources can only source or sink (absorb) power - not dissipate. There needs to be a resistive component (in the non-ideal source) for dissipation (losses) to occur.

Ideal Sources = dependent or independent ?

 

Doesn't matter if they are dependent or independent.

I think the issue is coming to grips with a source which absorbs or sinks power (really energy). The natural reaction is to ask where does the absorbed energy go. But you could equally well ask about a source supplying energy - "where did that energy come from?".

Sources used in circuit analysis are an ideal representation of reality - in the real world we have batteries and generators which either have stored energy or are mechanically driven to provide the energy flow - etc. Energy absorption in power supplies is not an unusual condition - consider the alternator in your car recharging the battery (electrical energy back to stored chemical energy) - or an electric train system which uses regenerative breaking (electrical energy back to mechanical).

 

i do understand the sources can consume , but the algebraic solution don't give zero , how can you explain that , this is the question .

thank you for your kind help

 

Maybe the attachment will help - I used a negative sign for both power lost in the resistors and absorbed by the 40V source. Both the dependent sources are supplying power to the circuit - hence the use of the positive sign. You didn't appear to have agreed with my comment about the output voltage signs for those two sources - they are in fact aiding each other - not opposing as you seem to think.

 

the 50 Ohm resistance is <?? Generating Power ??> , how ???

 

the 50 Ohm resistance is <?? Generating Power ??> , how ???

That would be something!

The resistances (50 Ω total) are dissipating power - power leaves the circuit
The 40 V source is absorbing power - power leaves the circuit
The two dependent sources are providing the (input) power which balances the power leaving the circuit.

 

Here is my effort:

\(40-(4V_1-V_2)+3V_3+V_2 = V_1+V_2+V_3\)

\(V_T = V_1+V_2+V_3\)

\(V_1=\frac{5}{50}V_T = 0.1V_T\)

\(V_2=\frac{25}{50}V_T= 0.5V_T\)

\(V_3=\frac{20}{50}V_T= 0.4V_T\)

\(40-(4V_1-V_2)+2V_3+V_2 = V_T\)

\(40-0.4V_T+0.5V_T+0.8V_T+0.5V_T = V_T\)

\(40+1.4V_T= V_T\)

\(1.4V_T= V_T-40\)

\(0.4V_T= -40\)

\(V_T= \frac{-40}{0.4}\)

\(V_T= -100\:Volts\)

................

\(V_1 = 0.1V_T = -10\)

\(V_2 = 0.5V_T = -50\)

\(V_3 = 0.4V_T = -40\)

................

\(4V_1-V_2= 4(-10)-(-50)=-40+50=10\)

\(2V_3+V_2=2(-40)+(-50)=(-80)+(-50)=-130\)

\(V_{(source)}= 40-10-130=40-140=-100\)

Provided my calculations are correct, the net input voltage (-100V) equals the total voltage drop across the 3 resistors (-100V).

Your result of -2 Amps times my net input voltage of -100 Volts yields 200 Watts input power. The I squared R power of each of the resistors summed together yields 200 Watts.

hgmjr

 

Here is my effort:

Provided my calculations are correct, the net input voltage (-100V) equals the total voltage drop across the 3 resistors (-100V).

Your result of -2 Amps times my net input voltage of -100 Volts yields 200 Watts input power. The I squared R power of each of the resistors summed together yields 200 Watts.

hgmjr

Hi hgmjr,

Your values are consistent with the OP's calcs. He also has I=-2A for the current - presumably assuming a clockwise convention in the equation setup.

I'm not 100% sure where CyborgBunny is having difficulties - it appears to be interpreting the power contributed by the three sources - one independent and two independent. I believe he wants to know why the total of the individual source and load (resistor) powers don't balance - by his reckoning. He also might have a solution from the original problem source which shows a deficit of 20W (?). One of his figures in the first post appear to have an error with respect to the resultant dependent source polarities. I haven't been able to draw a response from him on that point.

Perhaps you can elaborate for him when he reviews the latest posts.