EGR 2201 Unit 13 AC Power Analysis Read Alexander & Sadiku, Chapter 11. Homework #13 and Lab #13 due next week. Final Exam and Lab Exam next week.

Review: Power Supplies energy Absorb energy Recall the following key points about power from the first week of this course. An element’s power is the rate at which that element supplies or absorbs energy: Power’s unit of measure is the watt (W). By convention, we assign a positive sign to a power value if the element is absorbing energy, and we assign a negative sign if the element is supplying energy.

Review: The Power Law An element’s power is equal to the product of its voltage times its current: To get the correct sign (+ or ) on the power value when we use this equation, we must obey the passive sign convention, which says that we regard the positive direction for current as current into an element’s positive terminal.

Review: Dissipation versus Storage Recall also that resistors always absorb energy. They never supply energy. So a resistor’s power is always positive. The energy a resistor absorbs is lost (or “dissipated”) as heat. In contrast, inductors and capacitors are energy-storage elements. At times they may absorb energy, but at other times they may supply this energy back to the circuit. So an inductor’s or capacitor’s power may be positive at one time but negative at another time.

Review: Other Power Formulas for Resistors By combining the power law (p = v  i) with Ohm’s law (v = i  R or i = v  R), we can easily derive two other useful formulas for the power dissipated by a resistor: p = i 2  R p = v 2  R There are no similar formulas for capacitors or inductors in DC circuits. Do practice question.

Average Value of a Sinusoid (1 of 2) Consider a sinusoid that represents any quantity (voltage, current, power, …) versus time. If the sinusoid is symmetrical about the horizontal axis, then its average value is 0. In the circuits we’ve studied, a graph of voltage or current versus time would look like this. Therefore the average voltage or average current is 0.

Average Value of a Sinusoid (2 of 2) But if the sinusoid is “shifted up,” then its average value (see blue dashed line) is a positive number. As we’ll see, a graph of power versus time in an AC circuit would typically look like this. Therefore average power is usually not 0.

Shifting a Sinusoid Up Mathematically, we can shift a sinusoid up by adding a positive constant to the sinusoid. Example in MATLAB: >> fplot('5*cos(200*t)', [0, 0.1]) >> hold on >> fplot('3 + 5*cos(200*t)', [0, 0.1], 'r') What is the blue sinusoid’s average value? What is the red sinusoid’s average value? Do it in MATLAB, and then do practice question.

Average power (also called real power) Power in AC Circuits In AC circuits we distinguish several kinds of power: Quantity Symbol SI Unit Symbol for the Unit Instantaneous power p(t) watt W Average power (also called real power) P Apparent power S volt-ampere VA Complex power Reactive power Q volt-ampere reactive VAR

Instantaneous Power To find an element’s or network’s instantaneous power, use the same power formula as for DC circuits: The t’s remind us that in AC circuits, voltage and current change with time. So instantaneous power also changes with time. This equation holds whether the source is sinusoidal, triangle, square, etc. But we’ll focus on the sinusoidal case. Some of the equations in Chapter 11 are general, but many of them hold only for sinusoidal.

Multiplying Sinusoids In a network connected to a sinusoidal source, v(t) and i(t) are sinusoids with the same frequency. And p(t) = v(t) i(t), so p(t) is the product of two sinusoids. Question: What do you get when you multiply two sinusoids of the same frequency? Let’s use MATLAB to get an idea. Do it in MATLAB. >> fplot('5*cos(200*t)', [0, 0.1]) >> hold on >> fplot('8*cos(200*t+70*pi/180)', [0, 0.1],'r') >>fplot('5*cos(200*t)*8*cos(200*t+70*pi/180)',[0, .1],'g')

A Typical Graph of Instantaneous Power In typical AC circuits, a network absorbs energy during part of the cycle and supplies energy back to the source during part of the cycle. Therefore its power is sometimes positive and sometimes negative. Positive p(t): network is absorbing energy. Negative p(t): network is supplying energy.

Instantaneous Power with Sinusoidal Source Suppose a network’s voltage and current are 𝑣 𝑡 = 𝑉 𝑚 cos 𝜔𝑡+ 𝜃 𝑣 and 𝑖 𝑡 = 𝐼 𝑚 cos 𝜔𝑡+ 𝜃 𝑖 Then its instantaneous power is 𝑝 𝑡 = 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 + 1 2 𝑉 𝑚 𝐼 𝑚 cos 2𝜔𝑡+ 𝜃 𝑣 + 𝜃 𝑖 Do practice question. This term does not depend on t, and thus is constant. We call it the average power P. This term is a sinusoid whose frequency is twice the frequency of v(t) and i(t).

Graph of Instantaneous Power On the previous slide we had 𝑝 𝑡 = 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 + 1 2 𝑉 𝑚 𝐼 𝑚 cos 2𝜔𝑡+ 𝜃 𝑣 + 𝜃 𝑖 Constant term Sinusoid whose amplitude = 1 2 𝑉 𝑚 𝐼 𝑚 .

Average Power The constant term in our previous equation is the average power. It is measured in watts. 𝑝 𝑡 = 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 + 1 2 𝑉 𝑚 𝐼 𝑚 cos 2𝜔𝑡+ 𝜃 𝑣 + 𝜃 𝑖 So 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 . For any given network, cos 𝜃 𝑣 − 𝜃 𝑖 is a constant between 0 and 1, so P is a constant between 0 and 1 2 𝑉 𝑚 𝐼 𝑚 . Average power, P Do practice question.

Average Power is Real, Not Complex We have 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 . Note that everything on the right-hand side of this equation is real, not complex. Therefore, average power P always has a real value, not a complex value. So, for example, it would never be correct to write something like P = 4+j7 W or P = 830 W

Power Factor We have 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 . We call cos 𝜃 𝑣 − 𝜃 𝑖 the power factor. The symbol for power factor is pf. Its value is just a number, with no units. For any given network, pf is a constant between 0 and 1, so P is a constant between 0 and 1 2 𝑉 𝑚 𝐼 𝑚 . Power factor Do practice question.

Special Case #1: A Purely Resistive Network Recall that for a resistor or a resistive network, current and voltage are in phase with each other: 𝜃 𝑖 = 𝜃 𝑣 So the power factor is 1: cos 𝜃 𝑣 − 𝜃 𝑖 =1 And average power 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 simplifies to 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚

Other Average-Power Formulas for Resistors We’ve just seen that, for a resistor, 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 By combining this with Ohm’s law, we can easily derive two other useful formulas for the average power dissipated by a resistor: 𝑃= 1 2 𝐼 𝑚 2 𝑅 and 𝑃= 1 2 𝑉 𝑚 2 𝑅

Summary for Resistors DC AC 𝑝=𝑣𝑖 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 𝑝= 𝑖 2 𝑅 Compare the following formulas for computing a resistor’s power in a DC circuit and computing a resistor’s average power in a sinusoidal AC circuit: DC AC 𝑝=𝑣𝑖 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 𝑝= 𝑖 2 𝑅 𝑃= 1 2 𝐼 𝑚 2 𝑅 𝑝= 𝑣 2 𝑅 𝑃= 1 2 𝑉 𝑚 2 𝑅

Special Case #2: A Purely Inductive Network Recall that for an inductor or an inductive network, current lags voltage by 90: 𝜃 𝑖 = 𝜃 𝑣 −90° So the power factor is 0: cos 𝜃 𝑣 − 𝜃 𝑖 =0 And average power 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 simplifies to 𝑃=0

Special Case #3: A Purely Capacitive Network Recall that for a capacitor or a capacitive network, current leads voltage by 90: 𝜃 𝑖 = 𝜃 𝑣 +90° So the power factor is 0: cos 𝜃 𝑣 − 𝜃 𝑖 =0 And average power 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 simplifies to 𝑃=0

The General Case We’ve looked at three special cases: A purely resistive network: pf=1 and 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 = 1 2 𝐼 𝑚 2 𝑅. A purely inductive network: pf=0 and P=0. A purely capacitive network: pf=0 and P=0. In the general case, pf is a number between 0 and 1, and the formula for P does not simplify as it did in the special cases. We’re left with: 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖

The General Case In a general circuit containing sources, resistors, capacitors, and inductors, only the sources and the resistors have non-zero average power. The general formula 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 cos 𝜃 𝑣 − 𝜃 𝑖 applies to each element, but for the capacitors and inductors it simplifies to 0, and for the resistors it simplifies to 𝑃= 1 2 𝑉 𝑚 𝐼 𝑚 = 1 2 𝐼 𝑚 2 𝑅 Do practice question.

Review: Maximizing the Load Power In many applications, we wish to maximize the power transferred from a source to a load. Replacing the source with its Thevenin-equivalent circuit, we have the following situation: Thevenin-equivalent of source Variable load resistance

Review: The Load’s Power Depends on the Load Resistance For this circuit, the load resistor’s power is given by: 𝑝= 𝑖 2 𝑅 𝐿 = 𝑉 𝑇ℎ 𝑅 𝑇ℎ + 𝑅 𝐿 2 𝑅 𝐿 Question: For fixed values of VTh and RTh, what value of RL will result in maximum load power? The answer is not obvious, since RL appears in both the numerator and the denominator.

Review: Maximum Power Transfer Theorem For DC resistive circuits, the maximum power transfer theorem says that maximum power is transferred to a load when the load resistance equals the source’s Thevenin resistance (RL = RTh).

What About for AC Circuits? For AC circuits we have a similar situation, except instead of a Thevenin-equivalent resistance and a load resistance we have Thevenin-equivalent and load impedances.

Maximum Average Power Transfer Theorem for AC Circuits The maximum average power transfer theorem says that maximum power is transferred to a load when the load impedance equals the complex conjugate of the source’s Thevenin impedance: 𝐙 𝐿 = 𝐙 𝑇ℎ ∗ Also, 𝑃 𝑚𝑎𝑥 = 𝐕 𝑇ℎ 2 8 𝑅 𝑇ℎ , where 𝑅 𝑇ℎ is the real part of 𝐙 𝑇ℎ . Do practice question.

Different Ways to Give AC Values We’ve seen two ways to specify the size of an AC current or voltage: Peak-to-peak value Peak (or maximum) value A third way that is often used is called the effective value (or rms value). These distinctions apply only to AC, not to DC.

Can We Compare AC and DC? AC currents and DC currents are very different, but we can still draw some comparisons between them. For example: if an AC current flows through a resistor and a DC current flows through a resistor of the same size, each current will deliver power to its resistor.

The Idea Behind Effective Values For a given AC current, can we say what size DC current would deliver the same power to a resistor as the average power delivered by our AC current? Yes, we can, and we call the answer the effective value.

Effective Value So an AC current’s effective value is the DC current that delivers the same power to a resistor as the AC current delivers. An AC voltage’s effective value is defined in the same way.

Root-mean-square We’ve defined what we mean by effective value, but how can we compute effective values? Answer: to compute an AC current’s effective value, take the square root of the mean (average) of its square: 𝐼 𝑒𝑓𝑓 = 1 𝑇 0 𝑇 𝑖 2 𝑑𝑡 Effective values are also called rms (root-mean-square) values.

Root-mean-square for Sinusoids For a sinusoidal current, 𝑖 𝑡 = 𝐼 𝑚 cos 𝜔𝑡+ 𝜃 𝑖 , taking the root-mean-square is equivalent to dividing the current’s amplitude by 2 : 𝐼 𝑒𝑓𝑓 = 𝐼 𝑚 2 ≈0.707 𝐼 𝑚 Similarly for sinusoidal voltages. If 𝑣 𝑡 = 𝑉 𝑚 cos 𝜔𝑡+ 𝜃 𝑣 , then 𝑉 𝑒𝑓𝑓 = 𝑉 𝑚 2 ≈0.707 𝑉 𝑚 Do practice question.

Outlet Voltage in the USA The voltage at wall outlets in the USA is 120 V rms. This voltage is also a sinusoid, and it has a frequency of 60 Hz. Do practice question.

DC Versus AC on Multimeter Most digital multimeters can measure DC voltage, DC current, AC voltage, AC current. Voltage DC Voltage DC Current DC or AC? Current AC Voltage AC Current Tektronix CDM250 Fluke 45

DC or AC? When a multimeter is set to measure DC voltage or current, it actually displays the average value of the voltage or current. When a multimeter is set to measure AC voltage or current, it actually displays the rms (or effective) value of the voltage or current. Do practice question.