0:03

The topic of this problem is AC steady state power.

Â The problem is to determine the average power delivered to each of the elements in

Â the circuit shown below.

Â The circuit has two sources,

Â it has a current source which is a independent current source.

Â And it has a voltage source, which is a voltage controlled voltage source.

Â It's a dependent source where the voltage V sub 1, which is the voltage across

Â the 10 ohm resistor controls the overall voltage of this dependent voltage source.

Â We also have two resistors, a 10 and a 15 ohm resistor and

Â a capacitor with an impedance value of minus J20.

Â So if we're looking for

Â the average power delivered to the elements, we need to find the currents and

Â we need to find voltages so we can ultimately find the average power.

Â Remember that our equation for average power

Â is one-half V sub m, I sub m times a cosine of the angle

Â between those voltages and current levels.

Â 1:15

So perhaps one way to solve this problem is to find a nodal voltage at this point.

Â So use a nodal analysis summing currents into that node

Â to find the voltage at this point.

Â Once we know the voltage at this point,

Â we can find the current through the 10 ohm resistor.

Â We already know the current associated with the source so we have our Vs and

Â Is that we need in order to solve for our average power.

Â 2:06

And we're going to write a nodal equation summing the currents into node 1.

Â So we're going to sum the currents that are flowing into this node 1.

Â So first of all, we have the current which is from the source,

Â 20 at angle of 0 degrees is flowing into node 1.

Â We also have the voltage drop across a 10 ohm resistor which leads

Â to our current flowing into node 1.

Â And voltage drop is 0 for the ground node at the bottom minus V1 for

Â the voltage drop across the resistor divided by the resistance which is 10.

Â We also have this current which is flowing through the resistor and

Â capacitor back into node 1.

Â And that current is first of all the voltage drop from

Â one side which is minus three halves V1.

Â 3:09

On the right hand side of this series combination of resistor and

Â capacitor and then the voltage on the other side is V1.

Â So it's minus three-halves of V1 minus

Â V1 divided by 15 minus J20 for the impedance.

Â And the sum of those three currents at node 1 equal to 0 using

Â Kirkoff's Current Law.

Â 3:55

At an angle of minus 26.6 degrees.

Â That's our voltage at the nodal point one.

Â So if we have that voltage V sub 1,

Â then we can find everything else that we need in the current.

Â So if we're looking for, perhaps, this current I,

Â that's going through the capacitor and resistor on the top

Â right side of our circuit, then that current I is going to be V1

Â minus whatever the voltage is here, which is three-halves,

Â minus three-halves in V1 divided by the impedance which is 15 minus j20.

Â So the current I is going to be equal to V1.

Â Again, the voltage at node .1 minus a minus three-halves V1,

Â and minus three-halves V1 is the voltage at

Â the node on the right-hand side of our circuit,

Â divided by our Z value, 15 minus j20.

Â And so the current comes up to be five,

Â square-roots of five at an angle of 26.6 degrees.

Â That's for the current I that we've designated in our circuit.

Â So now we can use these voltages and

Â currents to find everything else we want to in the circuit.

Â 5:37

So the power average for the 10 ohm resistor is going to be one-half,

Â V sub m, I sub m cosine of the angle between them.

Â We know that since it's the resistor, the angle between the voltage and

Â current is 0, so we end up with a 1 for this cosine term.

Â So it's equal to one half V sub M,

Â I sub M and the cosine term is 1.

Â And so for the resistor, we have our V determined,

Â it's 50 square root of 5 at angle of minus 126.6 degrees.

Â So it's one-half, 50 square root of 5 for

Â our V sub m and then our I sub M is going to be 50 square roots of 5,

Â for the voltage drop, divided by 10 ohms.

Â And that's the current flowing downward in this case, through this 10 ohm resistor.

Â And as for the direction that we want to choose it based on our passive sign

Â convention where we've chosen the positive polarity of the voltage drop across

Â the 10 ohm resistor at the top of the resistor.

Â So through the passive sign convention we know that the current enters

Â the positive of the voltage drop across resistive or absorbing elements.

Â So defining this current is going to be V1 divided by ten.

Â And so V1 is 50 square root of 5, so

Â I'm going to square this term because we have two V sub ms and

Â we divide that by our R, which is 10.

Â So we end up with an average power absorbed by the resistor

Â which is 625 watts again, this is absorbed.

Â 7:51

We know that we're going to use the same equation that we had before for

Â the average power.

Â It's one half and the V sub m is the same voltage across

Â the current source as across the resistor and

Â it's equal to our nodal voltage, V sub 1.

Â So our V sub m is 50 square roots of 5.

Â Our I sub m is the current which is flowing down through

Â current source in this direction, since again,

Â we've chosen the voltage as plus to minus across the element.

Â So that current is minus 20, because we have a current source of

Â 20 at an angle of 0 going up through the left-hand leg of our circuit.

Â And then, we have the cosine of the angle between the voltage and

Â the current, the cosine of minus 26.6 degrees, and the current is 0 degrees.

Â So it's minus 26.6 minus 0

Â degrees for our angles.

Â And so we end up, for the current source,

Â we end up with a minus 1,000 watts power absorbed.

Â In other words, the current source is supplying 1,000 watts.

Â Now let's look at the 15 ohm resistor on the upper right-hand side of our circuit.

Â 9:37

So if we want to find that, then we have to use our current.

Â We know what the current I is.

Â The current I is 5 square roots of 5 at an angle of 26 degrees and

Â so we have our l sub n value.

Â We need to have a V sub m value as well.

Â Our V sub m value is going to be equal to the I times

Â the resistor because V is equal to IR.

Â And so for our voltage,

Â we have 5 square roots of 5 times 15,

Â that's our voltage and

Â our current is 5 square roots of 5.

Â So the power associated with the 15 ohm resistor and

Â we know that these angles are the same,

Â the angle between the current and the voltage are the same,

Â it's 26.6 for the voltage and 26.6 for the current.

Â 10:41

So for resistor again, that cosine is always equal to 1.

Â And so we end up with this 15 ohm resistor calculating

Â an average power absorbed by the resistor at

Â 937.5 watts again, absorbed power.

Â 11:18

Again we use our equation that we have above and we have a value for our current.

Â It's 5 square root of 5 and angle of 26.6 degrees and

Â we use the same approach as we did for the 15 ohm resistor.

Â We would take that current multiply it by our impedance value to get the voltage.

Â 11:41

We see that when we multiply the current, 5 square root 5 at an angle

Â of 26.6 degrees times this impedance which is 20 to angle minus 90 degrees.

Â Then we end up with a cosine which is between these

Â two angles theta V minus theta I of 90 degrees.

Â The difference in these two across associated with

Â this capacitor is 90 degrees, so the cosine is 0.

Â So the average power associated with the capacitor is 0 watts.

Â 13:06

And the average power for

Â that source is going to be one-half for

Â our term out front.

Â V sub m is going to be minus three-halves and

Â then we know what V1 is, 50 square root of 5.

Â So that's our VM term and then we have the current and

Â we also have our current term which is 5 square root of 5.

Â And we have the angle between those two,

Â cosine theta v which is minus 26.6

Â degrees minus theta sub i which is 26.6 degrees.

Â So our average power associated with

Â the dependent source is minus 562.5 watts.

Â And that's minus 562.5 watts absorbed power,

Â which means that the source is supplying 562.5 watts.

Â