# Won't the Power be dissipated in the first quadrant of this graph?

In the following picture: In the first quadrant of the V-I graph(for an Independent Current Source, where the voltage supplied is positive on the top) won't the power be dissipated instead of being generated?

However, the reference direction for $I_S$ is out of the positive labeled terminal, which is the active sign convention, and thus the current source absorbs power when the product of $v$ and $I_S$ is negative.
• @TejitPabari, if $I_S > 0$ and $v > 0$, the current source is supplying power to the external circuit, not absorbing power. This is easily verified by connecting a resistor to the terminals of the current source and observing that $v$ will be positive when $I_S$ is positive. Since the resistor absorbs power, it must be that the current source is supplying power. If the current source were absorbing power, the current through would be in to, not out of, the more positive terminal. – Alfred Centauri Jul 4 '16 at 21:07