In the following picture: Independent Current Source V-I graph

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?

Please Help! Thanks!

P.S. Independent Current Source = Where the current supplied is constant, despite any Voltage.


The picture is correct.

By the passive sign convention, the reference direction for current is into the positive labeled terminal of the circuit element and thus the circuit element is absorbs (not necessarily dissipates) power when the product of the voltage across and current through is positive.

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.

  • $\begingroup$ I am not sure I get you. Since the product is positive - the current is flowing in the same direction as the voltage (meaning from +ve to -ve) - shouldn't the power be absorbed ? However, when the current is in the opposite direction the power must be generated/ supplied, right? This is by the Passive Sign Convention. $\endgroup$ – Tejit Pabari Jul 4 '16 at 17:18
  • 1
    $\begingroup$ @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. $\endgroup$ – Alfred Centauri Jul 4 '16 at 21:07

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