ENSC2003 - Lecture 14

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ENSC2003 - Lecture 13
Power in DC Circuits - Slides

Power in Resistor-Only DC Circuits

$P=VI=\frac{V^2}{R}=I^{2}R$

• Resistors always consume power

Maximum Power Transfer Theorem

Given: voltage source $V_s$ with internal resistance $R_s$ + load $R_L$

• Max power over load occurs when:
  $R_L=R_s$
• Max power when condition is met:
  $P_{max}=\frac{V_s^2}{4R_s}$
• Max power dissipated by $R_s$ and $R_L$ are the same ($R_L=R_s$)
  • i.e. $R_L$ is using only half of the total power supplied

Maximum power transfer theorem shows a very low power efficiency (at most 50%) in DC resistor-only circuits

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Power in AC Circuits - Slides

Power in AC Circuits

• Instantaneous power $P(t)$ is a function of time in an AC circuit:
  $P(t)=v(t)i(t)$
  • Power is always greater than zero
• For a resistor:
  $P(t)=\frac{1}{R}{v}^2(t)=R{i}^2(t)$

Average and Peak Power

• Average power:
  $P_{avg}=\frac{V^2}{2R}$
• Peak power:
  $P_{pk}(t)=\frac{V^2}{R}$
  • Power of squared sinusoid: $P(t)=\frac{V^2}{R}{\cos }^2(\omega t+{\varphi }_v)$
  • Cosine term equals 1 or -1

Root Mean Square (RMS)

• Effective (DC-equivalent) value of AC:
  $V_{RMS}=\frac{V_{peak}}{\sqrt{2}}=\sqrt{\frac{1}{T}{\int }_{t=0}^T{v}^2(t)dt}$

• Power using RMS:
  $P_{avg}=\frac{V_{RMS}^2}{R}$

  • This is the main formula to be used in exams

• RMS current:
  $I_{RMS}=\frac{I}{\sqrt{2}}=\sqrt{\frac{1}{T}{\int }_{t=0}^T{i}^2(t)dt}$

Average voltage/current is constant; it does not depend on time

Power in Different Elements

• Capacitor: no net power consumed → $P_{avg}=0$
  • Stores energy in electric field but then returns it
• Inductor: no net power consumed → $P_{avg}=0$
  • Stores energy in magnetic field and then returns it

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Power in the AC Steady-State - Slides

Power in AC Steady-State

• Accounting for phase difference between $V$ and $I$:
  $P_{avg}=V_{RMS}{I}_{RMS}\cos ({\varphi }_v-{\varphi }_i)$

  • $\varphi ={\varphi }_v-{\varphi }_i$ → phase difference

• This equation above now works for all components:

  • Resistors work as normal as $I$ and $V$ are in phase
    • ${\varphi }_v-{\varphi }_i=0\to \cos ({\varphi }_v-{\varphi }_i)=1$
  • Capacitors and inductors consume $0$ power (see below)
    • ${\varphi }_v-{\varphi }_i=90°\to \cos ({\varphi }_v-{\varphi }_i)=0$
    • ${\varphi }_v-{\varphi }_i=-90°\to \cos ({\varphi }_v-{\varphi }_i)=0$

$V$ and $I$ are out of phase for capacitors and inductors: $P_{avg}=0$

• Capacitors/inductors do not dissipate real power, but they still reduce the real power dissipated by a resistor in the circuit
• This can be seen in equations when using:
  $\overline{I_{RMS}}=\frac{\overline{V_{RMS}}}{\bar{Z}}$
  • Capacitors/inductors affect impedance

Complex and Real Power

Complex power $S$ is defined as:
$S=V_{RMS}{I}_{RMS}\angle ({\varphi }_v-{\varphi }_i)=P+jQ$

• Has unit of Volt-Amperes (VA)
• $P$ is real component, $Q$ is imaginary component

Real power $P$ is the real component of $S$:
$P=V_{RMS}{I}_{RMS}\cos ({\varphi }_v-{\varphi }_i)$

• Has unit of Watt
• Only resistors can consume real power
• Real power $P$ is equal to $P_{avg}$
• Also known as active power
  $P=\sqrt{|S{|}^2-Q^2}$

Reactive power $Q$ is the imaginary component of $S$:
$Q=V_{RMS}{I}_{RMS}\sin ({\varphi }_v-{\varphi }_i)$

• Has unit of Volt-Amperes, Reactive (VAR)
• Only capacitors and inductors can consume reactive power
  $Q=\sqrt{|S{|}^2-P^2}$

Apparent Power $|S|$ is the magnitude of complex power $S$:
$|S|=V_{RMS}{I}_{RMS}$

• Has unit of Volt-Amperes (VA)
• Total “supplied” power

Power Factor

• Power triangle diagram:

  
  $\varphi ={\varphi }_v-{\varphi }_i$ → angle of power triangle

• If ${\varphi }_v<{\varphi }_i$, then $\varphi <0$ → current leads the voltage (leading)

• If ${\varphi }_v>{\varphi }_i$, then $\varphi >0$ → current lags the voltage (lagging)

The ratio of $P$ to $|S|$ is called the power factor, denoted by $pf$:
$pf=\frac{P}{|S|}=\cos \varphi =\cos ({\varphi }_v-{\varphi }_i)$

• Leading or lagging status must be indicated in $pf$ calculation
  • $\cos (\varphi )=\cos (-\varphi )$ -> we do not know if $Q$ is +ve or -ve
  • Thus we must indicate alongside $pf$ more information
• Low power factor → inefficiency
  • More current needed → greater power loss
• You want to maximise real component of complex power

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ENSC2003 - Lecture 15