Problem 1-15: Power of Independent Sources

Determine the power of each source.
a)
Voltage and Current Sources
b)
Voltage and Current Sources

Solution
a) The current source keeps the current of the loop  2A and the voltage source keeps the voltage across the current source  3v as shown below.

Voltage and Current Sources

For the voltage source the current enters from the positive terminal. Therefore, the passive sign convention should be used to find the sign of power:
 P=V \times I= 3 \times 2 =6W > 0 absorbing power
For the current source, the current leaves from the positive terminal. Hence, the active sign convention should be used:
 P= - V \times I= - 3 \times 2 =-6W < 0 supplying power

b) Similarily, the voltage across the current source and the currentpassing through thevoltage source can be easily determined by voltage and current sources, respectively:
Voltage and Current Sources
For this problem, the current enters to the voltage source from the negative terminal. Thus, the active sign convention should be used:

 P=-V \times I= -3 \times 2 =-6W  0 absorbing power

Comments

8 responses to “Problem 1-15: Power of Independent Sources”

  1. thirulok Avatar
    thirulok

    sir,
    whats the main difference of absorbing power and supplying power??
    mainly here, the circuits consists of 2 SOURCE….. which one is absorb the power from other???

    1. Dr. Yaz Z. Li Avatar

      Would you please clarify your question? In circuit a, the voltage source is absorbing current from the current source. In b, the current source is absorbing power from the voltage source.

      1. Hashem Avatar

        I think at the last line of the solution there is a mistake, you should complete the solving

        1. Yaz Avatar

          What is the mistake?

    2. salahuddin Avatar
      salahuddin

      if i m not wrong . Example, Absorbing power = when u connect source to battery, & supplying power = when u charge battery,

  2. Dhruv Avatar
    Dhruv

    i am totally confused when there is a voltage source and a current source in a circuit

    1. Mohamed Avatar
      Mohamed

      Me too

  3. […] used to find the sign of power: P_{V_S}=-V_S times I_1=-49.2 W < 0 (supplying power). Take a look here if you are would like to know how power of independent sources should be […]

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