# Electric Circuit Analysis: Calculating power absorbed/supplied by each element

• tahayassen
In summary, the textbook does not accurately explain another concept. The current coming out of the dependent source (p2) changes depending on the current through p1 branch (I), which is not constant. This causes confusion for students trying to calculate the power of a branch. Additionally, referencing the book properly would help receive more attention.
tahayassen
Once again, my textbook seems to do a poor job of explaining another concept just like I commented earlier here. Anyways, here is the relevant part of the textbook: http://www.mediafire.com/view/?cw5t2wr9vqqprnv

To calculate p4 in example 1.7, voltage of 8V is multiplied by a current of -0.2I. Firstly, why is the current negative? Secondly, why do they substitute a value of 5A for I. Where did they get this value?

I tried to do practice problem 1.7, but I don't seem to get the right answer for p3 when I use a voltage of 3V and a current of 0.6(3)A.

Because the dependent current source is supplying the power. So current flows from its positive to its negative. But when you calculate power of a branch you take the current value (which is to be multiplied) same as what will be entering the branch from positive.

They defined the dependent power that way. What ever will be the I (which is going through p1 brance) will be multiplied by 0.2 and that will be the current coming out of dependent source.

Kholdstare said:
Because the dependent current source is supplying the power. So current flows from its positive to its negative. But when you calculate power of a branch you take the current value (which is to be multiplied) same as what will be entering the branch from positive.

They defined the dependent power that way. What ever will be the I (which is going through p1 brance) will be multiplied by 0.2 and that will be the current coming out of dependent source.

Where did they get 5 from?

Well it says on your diagram I = 5 amps.

Incidentally, I cannot acces your link on my ofice computer, so others may well have the same problem.

It is better to post the doc or image here at PF, and also to properly reference the book as others may have it.

You will also likely receive more attention that way.

Studiot said:
Well it says on your diagram I = 5 amps.

Incidentally, I cannot acces your link on my ofice computer, so others may well have the same problem.

It is better to post the doc or image here at PF, and also to properly reference the book as others may have it.

You will also likely receive more attention that way.

I don't have time to do that at the moment as I'm heading to school right now, but isn't it 5 amps for only p2? For p3, it's 6 amps.

tahayassen said:
Where did they get 5 from?

The value of the second current source is not constant, but changes when the current through p2 branch changes (and the proportionality constant is 5).

tahayassen said:
I don't have time to do that at the moment as I'm heading to school right now, but isn't it 5 amps for only p2? For p3, it's 6 amps.

The I indicated in the figure defines the current through p2, not p3. Whatever is the value of current through p3 it will not necessarily be same as p2.

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Kholdstare said:

The value of the second current source is not constant, but changes when the current through p2 branch changes (and the proportionality constant is 5).
The I indicated in the figure defines the current through p2, not p3. Whatever is the value of current through p3 it will not necessarily be same as p2.

Ohh! I get it now! I feel so stupid. For some reason, I thought of I as the current flowing through p4 instead of thinking of I as a variable such as X that is clearly defined as 5A in the top left of the diagram. I'm so stupid.

Last edited by a moderator:

## 1. How do I calculate the power absorbed/supplied by each element in an electric circuit?

To calculate the power absorbed or supplied by an element in an electric circuit, you can use the formula P = VI, where P is power in watts, V is voltage in volts, and I is current in amps. Simply multiply the voltage and current values for each element to find the power. This formula applies to both DC and AC circuits.

## 2. Can I use Ohm's Law to calculate power in an electric circuit?

While Ohm's Law (V = IR) is a useful tool for calculating voltage, current, and resistance in a circuit, it cannot be used to directly calculate power. However, you can use Ohm's Law in combination with the power formula (P = VI) to find the power in a circuit.

## 3. How do I determine whether an element is absorbing or supplying power in a circuit?

An element in a circuit is absorbing power if the current enters the positive terminal and leaves through the negative terminal. On the other hand, an element is supplying power if the current enters through the negative terminal and leaves through the positive terminal. In other words, the direction of current flow determines whether an element is absorbing or supplying power.

## 4. Do I need to consider the power factor when calculating power in an AC circuit?

Yes, the power factor must be taken into account when calculating power in an AC circuit. The power factor is a measure of how efficiently the circuit is using the supplied power. It is calculated by dividing the real power (P) by the apparent power (S), where S = VI. The power factor ranges from 0 to 1, with 1 being the most efficient.

## 5. How do I calculate the total power in a series or parallel circuit?

In a series circuit, the total power is equal to the sum of the power for each individual element. In a parallel circuit, the total power is equal to the sum of the power for each branch of the circuit. To calculate the power for each branch, you can use the formula P = (V/R)^2, where V is the voltage and R is the resistance of the branch. Then, simply add the powers for each branch to find the total power in the parallel circuit.

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