And the 1 is the amplification factor for this current control voltage source.

Then if we look at the power in that case,

we see that the voltage across it is 1I sub x, minus to plus,

so it's going to be a -1I sub x for the voltage, and the current is 2 amps.

And again, that 2-amp current is encountering the negative polarity of

the dependent source first.

So it's (-1I sub x)(2A),

and so that gives us a -2 I sub x watts.

And we know what I sub x is, I sub x is equal to 4 amps.

So we can then plug that into this and we get a -8 watts

of power associated with that current control voltage source.

And as before, if we end up with a negative power,

then we know that that is supplied power.

So we have 8 watts of supplied power.