# Power through an unknown element

I've been stumped by a pretty simple question about power through an element in a circuit for a few days, and I was hoping someone could clear it up.

In the first diagram, the way I think of it is that the voltage from the bottom to the top, as indicated by the arrow and $$v_b$$, goes from negative to positive, and the current $$i_b$$ comes in through the positive side, so it's absorbing power. We use $$P=IV$$ and get an answer of $$30$$ watts absorbed by the element.

I employed the same logic for the second circuit, but I've been told repeatedly that my answer is wrong but I have no idea why. My logic is that the voltage between points D and E is $$20$$, and voltage goes from negative to positive, so D is the negative end and E is the positive end. So if a positive current of $$3$$ amps is flowing from E to D, then the current is entering from E, which is the positive end. So again, power is being absorbed, so $$P = IV$$ and we get $$60$$ watts absorbed. But everybody is telling me it's 60 watts supplied, or $$-60$$ watts absorbed.

Am I making a silly mistake or is my answer actually correct?

• You may be overthinking this. It really doesn't matter which end of the 2nd resistor is positive and which is negative. Resistors always dissipate energy (or "absorb" electrical energy and transform it into heat energy). The resistor is not going to supply electrical energy. – user93237 Sep 4 '19 at 21:36
• @SamuelWeir, why are you assuming they are resistors? – Alfred Centauri Sep 4 '19 at 21:39
• @AlfredCentauri - I guess because that's the international symbol for resistor and the poster wrote of "absorbing" power. – user93237 Sep 4 '19 at 22:39
• According to the counting conventions and the numbers these are NOT resistors – Hilmar Sep 5 '19 at 1:37

To me it is not clear as to the direction of the current $$i_{\rm ED}$$.
I have assumed that the rectangular boxes represent circuit elements rather than resistors?

Referring to the labels on the right hand diagram then for the left hand diagram if one labels the top node b and the bottom node a then $$v_{\rm b}$$ could be interpreted as $$v_{\rm ba}$$ which is the potential of node b relative to node a.
Adding an extra letter a to the suffix of the current $$i_{\rm b}$$ to give $$i_{\rm ba}$$ which is a current from node b to node a. Thus the current is flowing from node b to node a which in the given example is at a lower potential than node b.
The circuit element is thus a sink of electrical power, eg a resistor, with a numerical value for the power which is positive if one is using the passive sign convention.

For the right hand diagram the current is flowing from node E to node D which in the given example is at a higher potential than node E.
The circuit element is thus a source of electrical power, eg a battery, with a numerical value for the power which is negative if one is using the passive sign convention.

My logic is that the voltage between points D and E is 20, and voltage goes from negative to positive, so D is the negative end and E is the positive end.

But

$$v_{DE} = v_D - v_E = 20\,\mathrm{V}$$

thus

$$v_D > v_E$$

Put another way, the voltage $$v_{DE}$$ is the voltage measured by a voltmeter with the red lead on terminal $$D$$ and the black lead on terminal $$E$$.

So, if the voltmeter reading is positive, terminal $$D$$ is more positive than terminal $$E$$.

Double subscript notation

• Thank you so much! This bothered me so much for a while and it's finally off my mind :P – Lukasz Komza Sep 5 '19 at 2:17