# Why does this current source absorb but not deliver power?

Question (Extracted from a book): Determine the voltage across each current source, the current through each voltage source, and the power delivered or absorbed by each source.

Official Answer: Ix = 16 A, current through 20 V source is 6 A, voltage across VCCS is 12 V, and voltage across the CCVS is 8 V; 20 V source delivers 120 W; 10 A source delivers 200 W; VCCS absorbs 192 W; and CCVS absorbs 128 W.

Confusion:

10 A source delivers 200 W, 10V source delivers 200W, and CCVS absorbs 128 W. I have no problem with those, the current flows in the same direction as 10A, 20V and in the opposite direction of CCVS.

VCCS absorbs 192 W. This doesn't make sense. The current flows in the same direction as VCCS (0.8 Vab). It should deliver and not absorb power.

This seems like double-standard to me. The rules are: if the current flows from - to +, the source delivers power, and on the other side, if the current flows from + to -, the source is absorbing.

Why did the solution say the source is absorbing power and not delivering power?

• The 0.8 Vab current source confuses me too. Maybe it was a mistake and the arrow should point the other way. Ask the Prof.? Sep 5, 2014 at 21:45
• @GeorgeHerold If the arrow pointed the other way, it would be a completely different problem. Sep 5, 2014 at 21:52
• @GeorgeHerold, the VCCS has a transconductance of 0.8S so, if $V_{ab}$ is positive, the source drives a current 'down' equal to $V_{ab}\cdot 0.8S = 20V \cdot 0.8S = 16A$ Sep 5, 2014 at 22:05

VCCS absorbs 192 W. This doesn't make sense. The current flows in the same direction as VCCS (0.8 Vab). It should deliver and not absorb power.

But it does make sense since the source is absorbing power. According to the solution, the voltage across the VCCS is 12V with the top terminal more positive.

Since the 16A current enters the more positive terminal of the VCCS, power is delivered to the source. See the passive sign convention.

Note that, for the 10A source, the 10A current exits the more positive terminal so that source is delivering power.

Also, note that, for the 20V source, the 6A current exits the more positive terminal so that source is delivering power.

my main point of confusion is still why is VCSS + on top - on bottom and why is 10A - on top and + on bottom

The 10A source is + on top which is why it's delivering power. Look at the solution:

(1) due to the 20V voltage source on the left, the top circuit node is 20V more positive than the bottom node

(2) the center node between the two controlled sources must by 8V more positive than the bottom node (the CCVS is configured as a 1/2 ohm resistor and $16A \cdot 0.5 \Omega = 8V$)

(3) thus, by KVL, the top node must be 12V more positive than the center node.

• Note that, for the 10A source, the 6A current exits the more positive terminal. Why didn't the passive sign convention apply for the 10A source? You said that according to PSC, 16A current enters the more positive terminal of the VCCS; alright. With this logic, the 6A current should enter the positive terminal of 10A and exist the negative one.... Sep 5, 2014 at 22:31
• Be careful, that wiki page uses a different arrow convention for voltage: it doesn't show the direction of voltage drop, but the direction in which voltage grows. Sep 5, 2014 at 22:39
• @georgechalhoub, I've fixed the typo I introduced in an edit. My apologies if that confused you. Does it make more sense now? Sep 5, 2014 at 23:12
• @AlfredCentauri, the typo didn't confuse me. I hope you can still explain to me. Sep 5, 2014 at 23:14
• @AlfredCentauri my main point of confusion is still why is VCSS + on top - on bottom and why is 10A - on top and + on bottom Sep 5, 2014 at 23:27

If the voltage (drop) and the current across an element have the same direction (e.g. a resistor), the element absorbs power. On the other hand, if those two have different directions, the element delivers power. You are saying that "VCCS and $I_x$ have the same direction", but both of these are currents, so it can't be used to determine whether it absorbs or delivers power.

$V_{ab}=20V$, therefore $I_x=16A$ and the CCVS has a voltage of $8V$. $V_{VCCS}=V_{ab}-V_{CCVS}=20-8=12V$. So, the voltage of the VCCS drops "from top to bottom". That said, the voltage and current across it have the same direction, so it absorbs power ($16A\cdot 12V=192W$).

Calculations of the power absorbed/consumed by the devices:

20V source: 20V * (16A - 10A) = 120 Watts delivered (because current goes from - to + through the device)

10A source: 20V * 10 A = 200 Watts delivered (because current goes from - to + through the device)

CCVS: 8V * 16A = 128 Watts absorbed (because the current goes from the + to the - through the device)

VCCS: 12 * 16A = 192 Watts absorbed (because the current goes from the + to the - through the device)

The node between the CCVS and the VCCS is at 12Volts which is lower in potential than 20Volts at the top. This means that when you draw your + and - on the VCCS you need to make sure the + goes at the top and the - goes at the bottom.

• Why did the current went from + to - in VCCS, and it went from - to + in the 10A source? If the Passive Sign Convention made you draw the + the top and - to the bottom, then the Passive Sign Convention should make the current goes from + to - through the 10A source and not - to +. Sep 5, 2014 at 22:36
• @georgechalhoub The current went that way because if you solve the equations, that's the way the current went. The passive sign convention did not make me draw the +'s and -'s the way I did. The fact that the voltage is higher in certain areas is what determines how I labeled the signs on the components. Sep 5, 2014 at 23:30

I think there is some restrictions on dealing with ideal and dependant sources. If we are not careful, the circuit is contradictory and meaningless. For example, and ideal voltage source cannot be connected directly to another dependant voltage source if the voltages are not the same. Similarly, ideal current source cannot be connected directly to a dependant current source if the currents are not the same. The example above is a mix of two so that's the reason of confusion.