Reactance | PhysicsRank

An alternating current can average to zero and still heat a resistor. That is not a loophole in energy conservation: heating depends on the square of current. Phase and RMS values let us follow energy while the signs keep changing.

This chapter develops RMS from heating, phase from the element laws, and impedance from a phasor triangle. Each calculation keeps peak values, RMS values and average power separate.

What you will learn to do

Read amplitude, frequency and phase from a waveform.
Derive sinusoidal RMS and calculate RMS for pulses or offsets.
Obtain capacitor and inductor phase relations from their differential laws.
Solve series RLC current and component voltages using phasors.
Explain resonance and internal voltage magnification.
Derive average power and distinguish watts from apparent volt-amperes.
Relate power factor to source current and line losses.

Useful before this chapterInduction, moving conductors and magnetic energy →Potential, capacitors and dipoles →Oscillations: restoring forces, phase and energy →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

BEFORE YOU READ

Bring back one idea

Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.

Why does averaging the signed current first fail to predict resistor heating?

Compare your reasoning

Positive and negative current can cancel in the average, but each produces positive I²R heating. Average the square first, then take a square root to define the equivalent heating current.

Revisit the explanation →

01 / Alternating circuits: phase, RMS and real power

An alternating quantity needs an amplitude, frequency and phase

A sinusoidal voltage can be written v(t) = V0 cos(ωt + α). The amplitude V0 is its largest magnitude. The angular frequency ω = 2πf determines how quickly the cycle repeats, while α sets the position within the cycle at the chosen time origin.

One period T = 1/f corresponds to a phase advance of 2π radians. Changing the time origin changes the numerical phase but cannot change a measurable phase difference between two signals at the same frequency.

Steady-state sinusoidal analysis describes the response after switching transients have decayed. A real circuit just after a switch closes can contain both the driven sinusoid and a transient. Do not silently apply a steady-state phasor to an initial-condition problem.

Peak, instantaneous, average and RMS values answer different questions. Always identify which value a problem gives before using a circuit ratio or a power formula.

v(t)=V_0\cos(\omega t+\alpha),\qquad\omega=2\pi f,\qquad T=1/f

V0 is the peak magnitude, not automatically the meter reading or the cycle average.

WORKED EXAMPLE

Convert between angular frequency and period

A source has v(t) = 20 cos(100πt) V, with time in seconds. Find peak voltage, frequency and period.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: An alternating quantity needs an amplitude, frequency and phase
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
v(t)=V_0\cos(\omega t+\alpha),\qquad\omega=2\pi f,\qquad T=1/f
Meaning: V0 is the peak magnitude, not automatically the meter reading or the cycle average.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

02 / Alternating circuits: phase, RMS and real power

RMS is the constant value that gives the same resistive heating

A current that alternates symmetrically can have zero signed average while still heating a resistor. The instantaneous heating rate is i²R, which does not change sign when the current reverses. The useful average therefore involves the square of the current.

Define Irms = √⟨i²⟩, where the average is taken over a full period. Then the mean heating rate in a fixed resistor is Irms²R, exactly the rate from a constant current Irms. The same root-mean-square definition applies to voltage.

For a sinusoid, the average of cos² over a full cycle is one half, so Irms = I0/√2 and Vrms = V0/√2. The factor √2 is a waveform result, not the definition of RMS. A square wave, pulse train or distorted signal requires averaging its actual square.

An ordinary signed mean of a full-cycle sinusoid is zero. A mean absolute or rectified value is different again: for a sine wave it is 2I0/π. Neither signed mean nor rectified mean should be substituted for RMS in a heating calculation.

I_{\mathrm{rms}}=\sqrt{\frac1T\int_0^T i(t)^2\,dt},\qquad I_{\mathrm{rms}}=\frac{I_0}{\sqrt2}\ \text{for a sinusoid}

Squaring before averaging is essential. The RMS value is nonnegative.

WORKED EXAMPLE

Infer heating from a sinusoidal current

A 6.0 Ω resistor carries i(t) = 4.0 cos(ωt) A. Find RMS current and average heating power.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: RMS is the constant value that gives the same resistive heating
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
I_{\mathrm{rms}}=\sqrt{\frac1T\int_0^T i(t)^2\,dt},\qquad I_{\mathrm{rms}}=\frac{I_0}{\sqrt2}\ \text{for a sinusoid}
Meaning: Squaring before averaging is essential. The RMS value is nonnegative.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

03 / Alternating circuits: phase, RMS and real power

Compute RMS from the actual time fractions

For a waveform that stays at several fixed values, replace the integral by a time-weighted sum of squared values. Each squared level is multiplied by the fraction of the period spent there, then the square root is taken.

A symmetric square wave alternating between +I0 and −I0 has RMS I0 because its square is always I0². A pulse equal to I0 for fraction d of a period and zero otherwise has RMS I0√d, while its signed mean is I0d.

A sinusoid with a DC offset, i = Idc + I0 cos ωt, has mean square Idc² + I0²/2 because the cross term averages to zero. The DC and AC contributions add in mean square, not as RMS amplitudes.

These distinctions matter when a question changes a waveform shape while keeping its peak fixed. Equal peaks do not imply equal heating, and equal signed averages do not imply equal heating either.

WORKED EXAMPLE

Compare two signals that have the same signed average

One current is a constant 1.0 A. Another is 4.0 A for one quarter of each period and zero for the rest. Compare their mean currents, RMS currents and heating in the same fixed resistor.

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STOP & THINK

A symmetric square-wave voltage alternates between +12 V and −12 V. What is its RMS value?

0 V
12/√2 V
12 V

MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: Compute RMS from the actual time fractions
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

04 / Alternating circuits: phase, RMS and real power

Resistance gives an in-phase voltage and current

For an ideal resistor with constant R, vR = Ri at every instant. If current is I0 cos ωt, voltage is RI0 cos ωt. Both reach corresponding maxima and zeros together: their phase difference is zero.

The ratio of peak voltage to peak current equals R, and the ratio of RMS voltage to RMS current also equals R for the same waveform. Mixing a peak in the numerator with RMS in the denominator produces an artificial √2 factor.

Instantaneous absorbed power is Ri² and is never negative for positive R. Energy delivered to this element is converted to internal energy rather than periodically returned to the source in the ideal model.

Real components may contain additional inductance or capacitance, particularly at high frequency. The ideal resistor model here isolates the dissipative part so that phase relations can be built one element at a time.

V_{R,\mathrm{rms}}=RI_{\mathrm{rms}},\qquad P_R=I_{\mathrm{rms}}^2R

Voltage and current are in phase for an ideal fixed resistor.

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: Resistance gives an in-phase voltage and current
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
V_{R,\mathrm{rms}}=RI_{\mathrm{rms}},\qquad P_R=I_{\mathrm{rms}}^2R
Meaning: Voltage and current are in phase for an ideal fixed resistor.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

05 / Alternating circuits: phase, RMS and real power

A capacitor’s current follows the voltage’s rate of change

For an ideal capacitor, q = Cv and i = dq/dt = C dv/dt under the passive sign convention. If v = V0 cos ωt, then i = −ωCV0 sin ωt = ωCV0 cos(ωt + π/2). Current leads voltage by a quarter cycle.

This phase relation has a physical interpretation. At maximum capacitor voltage, stored charge is momentarily at an extremum, so current is zero. When voltage passes through zero most rapidly, charge changes most rapidly and current magnitude is largest.

The ratio of voltage amplitude to current amplitude is the capacitive reactance XC = 1/(ωC). It has units of ohms but does not describe resistive heating in an ideal capacitor. Higher frequency gives a larger rate of voltage change and therefore a larger current for the same voltage amplitude.

As frequency approaches zero, XC grows without bound. This is consistent with an ideal capacitor carrying no steady DC conduction current after it has charged. Switching transients must still be treated separately.

X_C=\frac1{\omega C},\qquad I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C},\qquad i\text{ leads }v_C\text{ by }\pi/2

The amplitude relation uses a sinusoidal steady state. An ideal capacitor stores and returns energy rather than dissipating it.

WORKED EXAMPLE

Use angular frequency directly and identify the phase

A 50 μF ideal capacitor is connected to a 20 V RMS sinusoidal source with angular frequency 200 rad/s. Find its reactance and RMS current.

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: A capacitor’s current follows the voltage’s rate of change
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
X_C=\frac1{\omega C},\qquad I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C},\qquad i\text{ leads }v_C\text{ by }\pi/2
Meaning: The amplitude relation uses a sinusoidal steady state. An ideal capacitor stores and returns energy rather than dissipating it.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

06 / Alternating circuits: phase, RMS and real power

An inductor’s voltage follows the current’s rate of change

For an ideal linear inductor in the passive sign convention, vL = L di/dt. If current is I0 cos ωt, voltage is −ωLI0 sin ωt = ωLI0 cos(ωt + π/2). Voltage leads current by a quarter cycle, or equivalently current lags voltage.

At maximum current, its derivative is zero, so the inductor voltage is zero. At the fastest change of current, the voltage magnitude is largest. This follows directly from the derivative and does not require a mnemonic.

The amplitude ratio is inductive reactance XL = ωL. Higher frequency requires a larger voltage to sustain the same current amplitude because the current must change more rapidly.

As frequency approaches zero, ideal inductive reactance approaches zero. A real coil still has winding resistance, and sudden switching may produce a substantial inductive voltage. The steady DC limit does not mean an inductor is always electrically invisible.

X_L=\omega L,\qquad V_{L,\mathrm{rms}}=X_LI_{\mathrm{rms}},\qquad v_L\text{ leads }i\text{ by }\pi/2

These expressions assume a linear ideal inductor in sinusoidal steady state.

WORKED EXAMPLE

Compare a coil at two frequencies

An ideal 0.30 H inductor carries 0.40 A RMS at angular frequency 100 rad/s. Find its RMS voltage. What voltage is needed for the same current amplitude if angular frequency doubles?

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Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: An inductor’s voltage follows the current’s rate of change
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
X_L=\omega L,\qquad V_{L,\mathrm{rms}}=X_LI_{\mathrm{rms}},\qquad v_L\text{ leads }i\text{ by }\pi/2
Meaning: These expressions assume a linear ideal inductor in sinusoidal steady state.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

07 / Alternating circuits: phase, RMS and real power

Add sinusoidal voltages with their phase relations intact

In a series RLC circuit the same current flows through all three elements. Use that current as the phase reference. The resistor voltage lies along the current phasor, the inductor voltage is 90° ahead, and the capacitor voltage is 90° behind.

The inductor and capacitor voltage phasors oppose one another. Their net transverse component is I(XL − XC). Combining it at right angles with IR gives source-voltage magnitude I√[R² + (XL − XC)²].

The impedance magnitude Z = √[R² + (XL − XC)²] relates matching RMS or peak source voltage and current. The source voltage leads current by φ with tan φ = (XL − XC)/R for positive R. Positive φ means an inductive circuit with lagging current; negative φ means a capacitive circuit with leading current.

Kirchhoff’s voltage law still holds at every instant. It is the scalar addition of separate peak or RMS magnitudes that generally fails, because those peaks occur at different times.

Z=\sqrt{R^2+(X_L-X_C)^2},\qquad\tan\phi=\frac{X_L-X_C}{R},\qquad I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}

Series RLC, sinusoidal steady state, positive constant R. Parallel circuits require adding branch currents or admittances instead.

WORKED EXAMPLE

A three-element circuit with a checkable phasor triangle

A series circuit has R = 30 Ω, XL = 70 Ω and XC = 30 Ω at the driving frequency. The source voltage is 100 V RMS. Find current, phase relation and RMS voltage across each element.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: Add sinusoidal voltages with their phase relations intact
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
Z=\sqrt{R^2+(X_L-X_C)^2},\qquad\tan\phi=\frac{X_L-X_C}{R},\qquad I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}
Meaning: Series RLC, sinusoidal steady state, positive constant R. Parallel circuits require adding branch currents or admittances instead.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

08 / Alternating circuits: phase, RMS and real power

Series resonance cancels reactance without erasing internal fields

In the ideal series model with fixed R, resonance occurs when XL = XC, giving ω0 = 1/√(LC). The source sees impedance magnitude R and current is in phase with source voltage. For fixed source amplitude, the current amplitude is then largest.

At resonance, the inductor and capacitor voltage phasors have equal magnitudes and opposite directions. Their sum is zero, but each can be large. The stored electric and magnetic energies continue to exchange during the cycle.

The ratio of either reactive voltage to source voltage at resonance is Q = ω0L/R = 1/(ω0CR). A large Q means large internal reactive voltages in this ideal series model. Real component losses and ratings matter in an actual circuit.

An ideal lossless series LC circuit driven exactly at resonance does not possess the finite steady-state amplitude predicted by dividing by R: R = 0 removes that steady-state assumption. Keep finite dissipation when discussing the usual resonance curve.

\omega_0=\frac1{\sqrt{LC}},\qquad Z(\omega_0)=R,\qquad Q=\frac{\omega_0L}{R}

These resonance statements apply to the stated series circuit with ideal L and C and frequency-independent positive R.

WORKED EXAMPLE

Find the resonance frequency and internal voltage magnification

A series circuit has L = 0.20 H, C = 20 μF and R = 10 Ω. It is driven at resonance by a 20 V RMS source. Find angular resonance frequency, current and each reactive RMS voltage.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: Series resonance cancels reactance without erasing internal fields
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
\omega_0=\frac1{\sqrt{LC}},\qquad Z(\omega_0)=R,\qquad Q=\frac{\omega_0L}{R}
Meaning: These resonance statements apply to the stated series circuit with ideal L and C and frequency-independent positive R.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

09 / Alternating circuits: phase, RMS and real power

Average power depends on both amplitude and phase

Let v = V0 cos ωt and i = I0 cos(ωt − φ). Multiply the two waveforms to obtain instantaneous power. The product identity gives a constant term V0I0 cos φ/2 and an oscillating term at twice the driving frequency.

Averaging over a full cycle removes the oscillating term. The result is Pavg = VrmsIrms cos φ. The factor cos φ is the displacement power factor for sinusoidal voltage and current.

An ideal capacitor or inductor has a phase difference of 90°, so its average absorbed power is zero even though instantaneous power alternates between positive and negative. Energy is stored during part of the cycle and returned during another part.

For a series RLC circuit, only R dissipates average power in the ideal model. Thus VrmsIrms cos φ must agree with Irms²R. This gives an independent check on impedance, phase and the use of RMS values.

p(t)=\frac{V_0I_0}{2}[\cos\phi+\cos(2\omega t-\phi)],\qquad P_{\mathrm{avg}}=V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi

The simple cosine power factor assumes sinusoidal signals at a common frequency. General distorted waveforms require averaging v(t)i(t).

Watch phase shift create intervals of returned energy

Inspect a plotted point: t / T = 0

Voltage / V₀: 1
Current / I₀: 0.5
Power / (V₀I₀): 0.5

Voltage is cos(2πt/T), current is cos(2πt/T − π/3), and normalized power is their product. Negative power means the load returns stored energy during that interval; the cycle average is still positive. Lines connect the plotted samples.

WORKED EXAMPLE

Check real power by two independent circuit relations

Use the earlier series circuit with R = 30 Ω, Z = 50 Ω, source 100 V RMS and current 2.0 A RMS. Find power factor and average power.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: Average power depends on both amplitude and phase
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
p(t)=\frac{V_0I_0}{2}[\cos\phi+\cos(2\omega t-\phi)],\qquad P_{\mathrm{avg}}=V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi
Meaning: The simple cosine power factor assumes sinusoidal signals at a common frequency. General distorted waveforms require averaging v(t)i(t).

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.

Back to chapter contents ↑

10 / Alternating circuits: phase, RMS and real power

Low power factor requires more current for a fixed useful power

At fixed RMS supply voltage and real power P, the required RMS current is P/(V cos φ) for a sinusoidal load. A smaller positive power factor therefore means a larger supply current.

If that current flows through a line resistance, line heating is I²Rline. Improving the power factor can reduce line losses while leaving useful load power unchanged. This is a comparison at fixed power and supply voltage, not at fixed impedance.

An inductive load can have its lagging reactive current partly balanced by a parallel capacitor’s leading current. Branch currents add as phasors; the source current decreases while the load’s own resistive demand can stay the same.

Power factor alone does not reveal whether current leads or lags: cos φ is even. The words leading or lagging, or the sign of reactive power, provide the missing direction. For nonsinusoidal loads, total power factor also reflects waveform distortion and need not equal a single phase cosine.

WORKED EXAMPLE

Compare line losses at two power factors

A load receives 600 W from a 100 V RMS sinusoidal supply. Compare supply current at power factors 0.60 and 1.00, and compare heating in the same 0.20 Ω line resistance. Treat the stated 100 V as the load-terminal voltage.

MEMBERSThe approach, step-by-step working and check are for members.Unlock →

Study with a tutor prompt

Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.

Topic: Alternating circuits: phase, RMS and real power
Focus: Low power factor requires more current for a fixed useful power
Learning goals:
- Read amplitude, frequency and phase from a waveform.
- Derive sinusoidal RMS and calculate RMS for pulses or offsets.
- Obtain capacitor and inductor phase relations from their differential laws.
- Solve series RLC current and component voltages using phasors.
- Explain resonance and internal voltage magnification.
- Derive average power and distinguish watts from apparent volt-amperes.
- Relate power factor to source current and line losses.
Equations in this section (LaTeX):
Build the physical reasoning before choosing an equation.

Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.

Back to chapter contents ↑

BRING THE IDEAS TOGETHER

What should stay with you

RMS is the square root of the mean square; the √2 rule belongs to sinusoids.
Current leads capacitor voltage and lags inductor voltage in sinusoidal steady state.
Series reactive voltages subtract as phasors, not as unsigned scalar magnitudes.
At series resonance source voltage and current align, while internal reactive voltages can remain large.
Average sinusoidal power is VrmsIrms cos φ and agrees with resistor heating.
Leading and lagging loads can have the same numerical power factor.

RETRIEVE, THEN EXPLAIN

Try it without the examples

Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.

SELF CHECK 1

A voltage is 10 V for 36% of a period and zero otherwise. What is its RMS value?

Answer in V.

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SELF CHECK 2

A series circuit has R = 12 Ω, XL = 25 Ω and XC = 9 Ω. What is its impedance magnitude?

Answer in Ω.

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SELF CHECK 3

A sinusoidal load takes 3 A RMS from 80 V RMS at power factor 0.50. What average power does it absorb?

Answer in W.

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SELF CHECK 4

At fixed sinusoidal voltage amplitude, increasing frequency across an ideal capacitor makes current amplitude:

Decrease
Increase
Stay fixed

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SELF CHECK 5

An ideal inductor has maximum positive current at a particular instant. What is its terminal voltage then in sinusoidal steady state?

Zero
Maximum positive
Maximum negative

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SELF CHECK 6

At series resonance with finite R, which statement is correct?

Both capacitor and inductor voltages must vanish
Their voltage phasors cancel while each may be large
The resistor dissipates no power

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SELF CHECK 7

Why is average absorbed power zero in an ideal sinusoidally driven capacitor?

It carries no current
It stores and returns equal energy over a cycle
Its instantaneous power is always zero

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APPLY THE CHAPTER

Take the ideas into a problem

Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.

Separate real power from impedanceAdvanced · Hints and guided reasoning The useful voltage measureIntermediate · Hints and guided reasoning A capacitor under alternating voltageIntermediate · Hints and guided reasoning

COME BACK TO THE IDEA

What can you explain later?

Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.

Next study session

Can a circuit have nonzero RMS voltage and current but zero mean real power in an ideal model?

Compare your reasoning

Yes. In an ideal purely reactive element, voltage and current are a quarter-cycle out of phase. Energy is exchanged with the source and returned over a cycle, giving zero mean real power.

Revisit the explanation →

A few days later: change the situation

May you divide every AC peak current by √2 to obtain RMS?

Compare your reasoning

No. That factor applies to a sinusoid. RMS must come from the mean square of the actual waveform; an ideal symmetric square wave has RMS equal to its peak magnitude.

Revisit the explanation →

Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.

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