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If you look at the fourth equation, the example is attempting to calculate the power supplied by the current-controlled current source on the far right of the circuit. For the equation P = I * V, they are plugging in 8 volts for V, which is the voltage across this current source resulting from the voltage source on the other side of the circuit.

What bothers me is that the current source should be outputting a voltage as well. Why is this not being factored in? Shouldn't the voltage value plugged into the power equation for that current source be equal to the 8 volts across it minus the voltage that it is outputting as a source?

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Since the current source is in parallel with p3, its voltage is the same as p3's: 8 volts. That is, 8 volts is "the voltage that it is outputting as a source". I'm not sure what other voltage you're talking about.

It's true that a real-world current source will include circuitry which will dissipate more power than is delivered to the load, but since there is no way to know what that is, even in principle, there is no way to calculate it and it is ignored. Just as the voltage source will, in real life, dissipate more power than it delivers to the load, that power is also ignored for the purposes of this sort of problem.

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  • \$\begingroup\$ Thanks for the help. One last clarification: we don't know what voltage the current source is outputting on its own, correct? we only know that the drop across it is 8V? \$\endgroup\$ – Sam D20 Sep 18 '15 at 20:09
  • \$\begingroup\$ @ByronS - Again; exactly. Just as an ideal voltage source puts out a voltage independent of current, a current source puts out current independent of the voltage needed. The range of available voltage for a real current source is called its compliance and is obviously not infinite. For problems like this, well, maybe it is infinite, but you don't need to worry about that, just like you don't need to worry about voltage sources supplying infinite current to a zero-ohm load. \$\endgroup\$ – WhatRoughBeast Sep 18 '15 at 20:14
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The calculation is correct, constant current source will provide constant current but voltage across it can vary based on the load.

Think of battery which is constant voltage source, it provides constant voltage but the current drawn through it can vary based on the load. In same manner, in constant current source, current is constant but voltage can vary based on load!

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Ohm's law states that only two of the three scalars (voltage, current, and resistance) are required to calculate the third, so there is no need to deal with the voltage at all since we can use its equivalent in the power formula.

\$P=I^2 R\$

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