only today , circuit help

please , can you help me with this question , i want to give it to the professor tommorow …

i have a solution , but not too sure … :frowning:

please help …

the q in the cloud are mine …


You need to work on LAYOUT!
It is very difficult to figure out anything - looks like a lot of hen scratchings and your writing is very difficult to read.

However, I will answer the easiest question first…
Right down the bottom you say “Does the total power supplied equal the total power absorbed…why?”
The answer to this is always YES, and the reason is that you cant get more energy out than you put in, and you cant put more energy in than you can get out… basic physics…

You say: “isnt *** and *** in parallel?” “why would they have different voltages?”

The answer is that they are not in parallel, they are in series! In Fig 0.0 they are in series parallel.

Call me crazy, but from my perspective, there is not enough information to give “the power absorbed by A” and “the power absorbed by each element” … maybe I am missing something?
The power is a function of the voltage and the current, but there are only voltages shown, so power cannot be calculated…

Unless of course this is all about algebra in which case I cant really help…

btw me = Advanced Trade Electrician

i managed to solve this question , by myself :slight_smile: , and it’s right , and yes , the power consumed came out to be the power generated , the conservation of power rule , i thought it’s a different case with current sources …

at the right , there’s a voltage controlled current source , and so i = 5 A , notice that if it was a voltage controlled voltage source then it would be -5 v no 5 v , allright … the rest it trivial …


thanks , i figured it out , and will see if there’s any oher comments , thanks .

Technically power absorbed and power supplied are equal in magnitude and opposite in “direction”, as it depends upon the circuit convention chosen (power is scalar, but read on).

P_abs + P_sup = 0 or P_abs = -P_sup

In active convention, we assume “electron” current or that current flows from negative to positive. In which case we draw a circuit device as (excuse the ASCII art ;)):

   |  +
 |   |     /\
 |   |     ||  i
   |   -

In passive convention (what most people are familiar with), we flip the direction of the current, i.e. we make it “conventional” current:

   |  +
 |   |     ||
 |   |     \/  i
   |   -

P = iv (in either case), So we see that the power supplied by the passive element and the active element have opposite signs. However, how can power be negative? An active element “supplies” power (or supplies energy to the circuit), and a passive element “absorbs” power (essentially dissipates energy from the circuit). So if we take the power supplied by a resistor (what we normally consider a “passive” element) it will be negative in the active convention (where an element supplies power). In the passive convention it will be positive (where each element absorbs power). Once again the sign of the power is only by convention, and for most circuit design applications all that really matters is the magnitude.

So what about P=(i^2)R and P=(v^2)/R, we obviously have positive magnitudes in both cases. Its important to apply the proper convention in these cases.

So for your specific problem you should get:

Power Absorbed by A (its in passive convention):
P = iv = (5*1)*9 = 45 W

For each element, the power supplied is:
For Vx, since its in active convention:

P = iv = (5*1)*1 = 5 W

For A, its in the passive convention, but we’re asked for its active convention, so current is flowing in the negative direction for the drawn polarity:

P = iv = (-5*1)*9 = -45 W

For the VCCS, its active convention:

P = iv = (5*1)*8 = 40 W

Just where in that diagram do you get a current value written at all that allows you to calculate power???
All I can see are lots of voltages…
Remembering that
V = Voltage
I = Current

I am really interested to know.

There is a VCCS (voltage controlled current source) (see this).

Since the current in the loop is constant, and given to be 5Vx, which we’re given Vx = 1, then we know current.

Glad you received lots of help 3D “g”. Your prof will be impressed no doubt.

Kia Ora Bro.