Solving for the Average Power and the Reactive Power

Thread Starter

omarsinnno

Joined Feb 5, 2019
8
In the following exercise they asked to find the average power of the 5 ohm resistor, then the average power of the 8 ohm resistor, and then deduce the the voltage source (rms) value.

Given
: -2j capacitor absorbs -2.3 VARs.

I did the first part and found out that the average power of the 5 ohm resistor is 0.9199 W by finding that the Vrms = 3.033150178 V.

I haven't successfully used the information given or deduced so far to find the average power in the 8 Ohm resistor.
Any help is really appreciated!

 
Last edited:

WBahn

Joined Mar 31, 2012
24,849
It's the same as the total current in the circuit.
Think about that statement for a moment. That junction has four branches, so what does it even mean to say that it's the same as a specific current?

What does KCL say about the current at that node?
 

Thread Starter

omarsinnno

Joined Feb 5, 2019
8
Think about that statement for a moment. That junction has four branches, so what does it even mean to say that it's the same as a specific current?

What does KCL say about the current at that node?
The sum of all currents at that node should be equal to zero. In other words, the current entering the node is equal to the one leaving.
 

WBahn

Joined Mar 31, 2012
24,849
The sum of all currents at that node should be equal to zero. In other words, the current entering the node is equal to the one leaving.
Okay. So can you define all of the currents entering and leaving that node and set up an equation that applies that constraint?

Next, what can you say about the voltage across some of the components relative to the others?

You could also approach this from the opposite direction. Simply solve the entire circuit in terms of Vs, including the power in each component. The use the known power in the capacitor to find Vs and use Vs to answer the rest of the questions.

Even better practice would be to solve for all of the requested quantities in terms of the power in the capacitor and then plug that known power into each one in turn.
 

The Electrician

Joined Oct 9, 2007
2,750
I did the first part and found out that the average power of the 5 ohm resistor is 0.9199 W by finding that the Vrms = 3.033150178 V.
This "Vrms" you refer to, is it the voltage (with respect to the "minus" terminal of Vs) at the node where the 8 ohm and 5 ohm resistors are connected? Please show how you determined this value.
 

Thread Starter

omarsinnno

Joined Feb 5, 2019
8
Knowing that in a capacitor P = 0, then the reactive factor is equal to 1 giving us:

(Vm.Im)/2 = 2.3

Im = Vm/Z

We then get Vm^2 = 9.2, giving me the answer I eventually got to for Vm.
 
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