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I had a question during a physics tutorial today and it has left me quite confused.

Here's a picture of the circuit diagram

One of the questions was to find the power delivered by the 6V source. I figured that P=VI, I is negative with respect to the voltage source, so therefore power is negative. So the voltage source must be absorbing power?

This didn't make much sense, and my tutor assured me that voltage sources can only supply power, and that I should just forget about the negative sign. Is this correct, or is there something wrong with the circuit?

Thank you for any feedback.

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  • \$\begingroup\$ Did the teacher happen to give you a value for for the unspecified voltage source below the \$120\Omega\$ resistor? And voltage sources can both supply power and absorb power. (But not simultaneously.) \$\endgroup\$
    – jonk
    Sep 3, 2016 at 3:15
  • \$\begingroup\$ @jonk about 6.527V by a rough calculation. \$\endgroup\$
    – Tom Carpenter
    Sep 3, 2016 at 3:21
  • \$\begingroup\$ @TomCarpenter: My brain went somewhere else with the 'A'. It's an ammeter, then. \$\endgroup\$
    – jonk
    Sep 3, 2016 at 3:26
  • \$\begingroup\$ @jonk the first time I looked at the circuit, I read current source (its the arrow what confused me) until I looked again. \$\endgroup\$
    – Tom Carpenter
    Sep 3, 2016 at 3:33
  • \$\begingroup\$ @TomCarpenter: Yeah. Sadly, I didn't spend time looking again and needed your clue. As soon as I read your comment, it all came together of course. But I should have taken the same moment you did. Oh, well. I'll learn. \$\endgroup\$
    – jonk
    Sep 3, 2016 at 3:34

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An ideal voltage source can supply or absorb current, which is determined by the circuit. The only rule of an ideal voltage source is that the terminal voltage is always the specified value, no matter what the current is.

This basically means that either your tutor or your understanding of your tutor is incorrect as voltage sources can effectively act as both a supplier of power and a consumer of power - which it is depends on the rest of the circuit.

In your circuit, two of the voltage sources are in fact acting as loads. Only the 12V source is supplying power.

In practice an ideal voltage source is never achievable. There are no known practical voltage sources that can supply infinite current and as such there are no ideal voltage sources. This is usually described by adding a resistor in series with the voltage source to represent the "internal resistance" of the supply - or in other words to approximate the limited current sourcing capability of the supply.

As a point of interest, an ideal voltage source which is absorbing power is actually equivalent to an ideal diode - the forward voltage of the diode is represented by the supply voltage.

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  • \$\begingroup\$ Thanks for the comprehensive response, you confirmed what I already thought. I've described voltage sources as consuming power in my other electrical courses and never had a problem. It's only because my physics tutor said the opposite that I wanted to double check my understanding. Just confirming though, the correct answer to the question: "What is the power supplied by the 6V source?" should be -72mW. Don't worry, I'm not going to be a dick about it to my tutor, I just want to make sure I know the correct answer. Cheers. \$\endgroup\$
    – Tom Bailey
    Sep 3, 2016 at 3:58
  • \$\begingroup\$ @TomBailey Technically yes, it is -72mW - in the sense that a negative value indicates power being absorbed. \$\endgroup\$
    – Tom Carpenter
    Sep 3, 2016 at 4:02

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