Physics Formulae Electricity Ohm's Law Of AC Current containing Only Resistance R

Subset – Definition and Properties

Learn Ohm's Law for AC current in purely resistive circuits. This formula calculates voltage or current, which remain pe...

Ohm's Law in AC Circuits (Pure Resistance)

Ohm's Law in AC circuits with pure resistance maintains the same simple linear relationship as in DC circuits: voltage equals current times resistance (V = IR). The key insight is that pure resistors do not introduce any phase shift between voltage and current — they remain perfectly in phase at all frequencies. This means both voltage and current reach their maximum and minimum values simultaneously, and both cross zero at the same instants. The resistance value R remains constant regardless of AC frequency, making resistive circuits the simplest case in AC analysis. This fundamental relationship applies whether using RMS values for power calculations or instantaneous values for time-domain analysis.

Physically, pure resistance in AC circuits represents the direct conversion of electrical energy into heat without any energy storage. Unlike reactive components (inductors and capacitors), resistors respond instantaneously to voltage changes. The electrons flowing through the resistive material experience friction-like collisions that convert kinetic energy to thermal energy, regardless of whether the current is DC or AC. This instantaneous response means no phase shift occurs. The frequency independence arises because the microscopic resistance mechanism (electron-lattice interactions) operates much faster than typical AC frequencies.

Physical Properties

Ohm's Law for an AC circuit containing only a pure resistor describes a simple, linear relationship between voltage, current, and resistance, identical to its DC counterpart. The key physical property is that the resistor does not introduce any time delay, or phase shift, between the voltage across it and the current flowing through it; they remain perfectly synchronized.

PropertyDetails
NatureVoltage (V), Current (I), and Resistance (R) are treated as scalar quantities for magnitude calculations (e.g., RMS or peak values).
SI UnitsVoltage: Volt (V), Current: Ampere (A), Resistance: Ohm (Ω)
Magnitude RelationshipThe RMS voltage is directly proportional to the RMS current: V_rms = I_rms * R. The same holds for peak values: V_peak = I_peak * R.
Phase RelationshipIn a purely resistive AC circuit, the voltage and current are always in phase. The phase angle difference between them is 0 degrees.
Conservation LawRepresents conservation of energy, where electrical energy is dissipated as heat in the resistor. The average power dissipated is P = (I_rms)^2 * R.
Dimensional FormulaThe dimensional formula for Resistance (R) is [M L^2 T^-3 I^-2].
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Diagram & Visualization

V(t) R I(t) 0 t V(t) I(t)
In a purely resistive AC circuit, the voltage (V) and current (I) waveforms remain perfectly in phase.
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Key Formulas

\[ u(t) = i(t) R \]
Instantaneous Form
\[ U_{\text{rms}} = I_{\text{rms}} R \]
RMS Form
\[ U_0 = I_0 R \]
Peak (Amplitude) Form
\[ P_{\text{avg}} = U_{\text{rms}} I_{\text{rms}} = \frac{U_{\text{rms}}^2}{R} = I_{\text{rms}}^2 R \]
Average Power Dissipation
\[ p(t) = u(t) i(t) = U_0 I_0 \cos^2(\omega t) \]
Instantaneous Power
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Variables

SymbolQuantitySI UnitDescription
\(u(t), v(t)\)Instantaneous VoltageVolt (V)The voltage at a specific moment in time.
\(i(t)\)Instantaneous CurrentAmpere (A)The current at a specific moment in time.
\(U_{\text{rms}}, V_{\text{rms}}\)RMS VoltageVolt (V)The Root Mean Square or effective value of the AC voltage.
\(I_{\text{rms}}\)RMS CurrentAmpere (A)The Root Mean Square or effective value of the AC current.
\(U_0, V_0\)Peak VoltageVolt (V)The maximum amplitude of the AC voltage waveform.
\(I_0\)Peak CurrentAmpere (A)The maximum amplitude of the AC current waveform.
\(R\)ResistanceOhm (Ω)The opposition to current flow, independent of frequency.
\(P_{\text{avg}}, P\)Average PowerWatt (W)The average rate of energy conversion to heat over a full cycle.
\(p(t)\)Instantaneous PowerWatt (W)The power at a specific moment in time; always positive for a resistor.
\(\phi\)Phase AngleRadian (rad) or Degree (°)The phase difference between voltage and current. For a pure resistor, \(\phi = 0\).
\(\cos(\phi)\)Power FactorDimensionlessThe ratio of real power to apparent power. For a pure resistor, \(\cos(\phi) = 1\).
\(Z\)ImpedanceOhm (Ω)The total opposition to AC current. For a pure resistor, \(Z = R\).
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Derivation of Zero Phase Shift

We can prove that the voltage and current are in phase in a purely resistive AC circuit by starting with the general form of an AC voltage and applying Ohm's law.

1. Assume a sinusoidal voltage source is applied across a resistor R. The instantaneous voltage \(u(t)\) can be described by:

\[ u(t) = U_0 \cos(\omega t + \phi_u) \]
General AC Voltage

where \(U_0\) is the peak voltage, \(\omega\) is the angular frequency, and \(\phi_u\) is the initial phase angle of the voltage.

2. According to Ohm's law for instantaneous values, the current \(i(t)\) is directly proportional to the voltage \(u(t)\) at every instant:

\[ i(t) = \frac{u(t)}{R} = \frac{U_0 \cos(\omega t + \phi_u)}{R} \]

3. We can define the peak current \(I_0\) as \(U_0 / R\). Substituting this into the equation gives the expression for instantaneous current:

\[ i(t) = I_0 \cos(\omega t + \phi_u) \]

4. The general form for the current is \(i(t) = I_0 \cos(\omega t + \phi_i)\). By comparing this with our derived expression, we see that the phase angle of the current, \(\phi_i\), is identical to the phase angle of the voltage, \(\phi_u\).

\[ \phi_i = \phi_u \]

5. The phase difference, \(\phi\), between voltage and current is defined as \(\phi = \phi_u - \phi_i\). Therefore:

\[ \phi = \phi_u - \phi_i = 0 \]
Zero Phase Shift

This proves that for a purely resistive AC circuit, the voltage and current waveforms are perfectly in phase.

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Types & Special Cases

While the fundamental law V = IR remains constant, its application in AC circuits can be classified based on the type of values being used to represent the time-varying voltage and current.

Type / CaseDescriptionWhen to Use
Instantaneous FormRelates the voltage v(t) and current i(t) at any specific moment in time. For example, if i(t) = I_peak * sin(ωt), then v(t) = (I_peak * R) * sin(ωt).Useful for analyzing waveform behavior and understanding the in-phase relationship at any point in the cycle.
Peak Value FormRelates the maximum amplitude of the voltage and current waveforms: V_peak = I_peak * R.Used in applications where the maximum voltage or current is a critical design parameter, such as determining component ratings to prevent damage.
RMS (Root Mean Square) FormRelates the effective values of AC voltage and current: V_rms = I_rms * R. RMS values provide the equivalent DC values for power dissipation.This is the most common form for general AC circuit analysis, power calculations, and when using standard AC measuring instruments (voltmeters, ammeters).

Worked Example

Given an AC circuit with a resistor of \(R = 60 \) Ω and an RMS voltage of \(V_{\text{rms}} = 240 \) V, find the RMS current \(I_{\text{rms}}\) and the average power \(P_{\text{avg}}\) dissipated by the resistor.
  1. Apply Ohm's law for AC circuits using RMS values to find the current: \(V_{\text{rms}} = I_{\text{rms}} R\).
  2. Rearrange the formula to solve for \(I_{\text{rms}}\): \(I_{\text{rms}} = V_{\text{rms}} / R\).
  3. Substitute the given values: \(I_{\text{rms}} = 240 \text{ V} / 60 \text{ Ω}\).
  4. Calculate the RMS current.
  5. Use the power formula \(P_{\text{avg}} = V_{\text{rms}} \times I_{\text{rms}}\) to find the average power.
  6. Substitute the known values and calculate the power.
The RMS current is \(I_{\text{rms}} = 4 \) A. The average power dissipated is \(P_{\text{avg}} = 240 \text{ V} \times 4 \text{ A} = 960 \) W.
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Try It

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Applications

Heating Elements: Devices like electric stoves, water heaters, and space heaters use resistive elements to efficiently convert electrical energy into thermal energy. Their behavior is almost purely resistive, with a unity power factor.

Lighting Systems: Incandescent and halogen bulbs operate by passing current through a tungsten filament, which acts as a resistor. The heat generated causes the filament to glow, producing light.

Industrial Processes: High-power resistive heating is used in electric furnaces for melting metals, glass manufacturing, and in welding applications where precise and intense heat is required.

Electronic Circuits: Resistors are fundamental components used for current limiting, voltage division, signal conditioning, and as pull-up/pull-down resistors in digital logic circuits. Their predictable, frequency-independent behavior is crucial for circuit design.

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Real-World Numerical Examples

A standard US wall outlet provides an RMS voltage of 120 V at 60 Hz. If a simple electric toaster with a heating element resistance of 10 Ω is plugged in, what is the RMS current drawn by the toaster and how much power does it consume?
  1. Identify the given values: \(V_{\text{rms}} = 120 \) V, \(R = 10 \) Ω.
  2. Use Ohm's Law for AC RMS values to find the current: \(I_{\text{rms}} = V_{\text{rms}} / R\).
  3. Calculate the current: \(I_{\text{rms}} = 120 \text{ V} / 10 \text{ Ω} = 12 \) A.
  4. Calculate the average power using the formula \(P = V_{\text{rms}} \times I_{\text{rms}}\).
  5. Substitute the values: \(P = 120 \text{ V} \times 12 \text{ A} = 1440 \) W.
  6. Alternatively, use \(P = V_{\text{rms}}^2 / R = (120)^2 / 10 = 14400 / 10 = 1440 \) W.
The toaster draws an RMS current of 12 A and consumes 1440 W of power. The voltage and current are in phase.
An electronics technician is testing a precision 50 Ω power resistor. They apply three different AC signals: 1) 120 V RMS at 60 Hz, 2) 10 V RMS at 1 kHz, and 3) 5 V RMS at 1 MHz. Verify that the resistance is consistent across these frequencies by calculating the current and power for each case.
  1. <strong>Case 1 (60 Hz):</strong> Calculate current \(I_1 = V_1/R = 120/50 = 2.4 \) A. Calculate power \(P_1 = V_1 I_1 = 120 \times 2.4 = 288 \) W.
  2. <strong>Case 2 (1 kHz):</strong> Calculate current \(I_2 = V_2/R = 10/50 = 0.2 \) A. Calculate power \(P_2 = V_2 I_2 = 10 \times 0.2 = 2 \) W.
  3. <strong>Case 3 (1 MHz):</strong> Calculate current \(I_3 = V_3/R = 5/50 = 0.1 \) A. Calculate power \(P_3 = V_3 I_3 = 5 \times 0.1 = 0.5 \) W.
  4. <strong>Verification:</strong> For each case, check if \(R = V/I\) holds. \(R_1 = 120/2.4 = 50\) Ω. \(R_2 = 10/0.2 = 50\) Ω. \(R_3 = 5/0.1 = 50\) Ω.
The calculated resistance is 50 Ω in all three cases, demonstrating the frequency-independent nature of an ideal resistor. The currents are 2.4 A, 0.2 A, and 0.1 A, and the power dissipations are 288 W, 2 W, and 0.5 W, respectively.
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Real-World Scenarios

V , I
Electric Heater
An electric heater's resistive element converts AC electrical energy directly into heat, with voltage and current remaining in phase.
Incandescent Bulb
The filament in an incandescent bulb acts as a resistor, heating up and glowing when AC current flows through it, demonstrating P = V²/R.
In Out
Volume Control
A potentiometer acts as a variable resistor, attenuating an AC audio signal to control the sound level by changing the resistance in the circuit.

Electric Heaters
Space heaters, electric stoves, and toasters are common household appliances that function as nearly pure resistive loads. When connected to an AC outlet, they draw a current that is in phase with the voltage, and their primary purpose is to dissipate this electrical energy as heat according to the formula \(P = I^2R\).

Incandescent Light Bulbs
Though largely replaced by newer technologies, the classic incandescent bulb is a perfect example of a resistive load. A thin tungsten filament resists the flow of AC current, heating up to the point of incandescence and emitting light. The brightness can be controlled by varying the voltage with a dimmer, which directly alters the power dissipated (\(P = V^2/R\)).

Volume Control Knobs
In older audio equipment, a variable resistor (potentiometer) is used as a voltage divider to control volume. As you turn the knob, you change the resistance in the path of the audio signal (a complex AC waveform), which attenuates the voltage delivered to the amplifier, thereby adjusting the sound level.

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Limitations and Assumptions

⚠️ The assumption that resistance is independent of frequency is only valid for ideal resistors. Real-world resistors exhibit parasitic effects at very high frequencies.

At high frequencies (typically in the MHz to GHz range), a physical resistor begins to behave like an RLC circuit due to:

  • Parasitic Inductance (L): The leads and body of the resistor form a small inductor.
  • Parasitic Capacitance (C): There is a small capacitance between the resistor's terminals.

The actual impedance (Z) becomes frequency-dependent:

\[ Z_{\text{actual}} = R + j\omega L_{\text{parasitic}} + \frac{1}{j\omega C_{\text{parasitic}}} \]
💡 For most common applications, including household power (50/60 Hz) and audio frequencies, these parasitic effects are negligible and resistors can be treated as ideal (Z ≈ R).

Common Mistakes

⚠️ Confusing Peak vs. RMS Values: Using peak voltage (\(V_0\)) or peak current (\(I_0\)) in power formulas that require RMS values. For example, calculating average power as \(P = V_0 I_0\) is incorrect; the correct formula is \(P = V_{\text{rms}} I_{\text{rms}} = (V_0 I_0)/2\).
⚠️ Assuming Resistance Changes with Frequency: Mistakenly thinking that since reactance of capacitors and inductors changes with frequency, the resistance of a resistor must also change. For an ideal resistor, R is constant across all frequencies.
⚠️ Ignoring Phase: While the phase angle is zero for a pure resistor, students sometimes forget this key property when moving to more complex RLC circuits. The resistor is the only component where voltage and current are in phase, serving as the reference for phase analysis.
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Units and Dimensions

QuantitySymbolSI UnitDimensional Formula
Voltage\(V, U\)Volt (V)\([M L^2 T^{-3} I^{-1}]\)
Current\(I\)Ampere (A)\([I]\)
Resistance\(R\)Ohm (Ω)\([M L^2 T^{-3} I^{-2}]\)
Impedance\(Z\)Ohm (Ω)\([M L^2 T^{-3} I^{-2}]\)
Power\(P\)Watt (W)\([M L^2 T^{-3}]\)
Frequency\(f\)Hertz (Hz)\([T^{-1}]\)
Angular Frequency\(\omega\)radians/second (rad/s)\([T^{-1}]\)
Phase Angle\(\phi\)Radian (rad)Dimensionless
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Study Strategy

1 🧠 Grasp the Fundamentals
  • Read the DEFINITION section, focusing on why voltage and current are perfectly in phase in a purely resistive AC circuit.
  • Contrast the AC formula with DC Ohm's Law. Note that V and I now represent RMS or peak values, not constant DC values.
  • Understand that for a pure resistor, the resistance value R is constant and does not change with the AC frequency.
  • Draw a simple circuit diagram of an AC voltage source connected to a single resistor, labeling V(t), I(t), and R.
2 📝 Commit the Formula to Memory
  • Write down the three forms of the law: V_rms = I_rms * R, V_peak = I_peak * R, and v(t) = i(t) * R.
  • Sketch the phasor diagram for a resistor, showing the voltage and current phasors aligned with a zero-degree phase angle.
  • Draw the sinusoidal waveforms for voltage and current on the same axes, showing they reach their peaks and zeros simultaneously.
  • Create a concept map linking 'pure resistor' to 'in phase', 'V=IR', and 'unity power factor'.
3 ✍️ Practice with Problems
  • Work through the Worked Example problems step-by-step, ensuring you can calculate both RMS and peak values correctly.
  • Review the COMMON_MISTAKES section. Create a problem that forces you to distinguish between V_peak and V_rms when calculating average power.
  • Find a problem where you are given RMS voltage and must find peak current. This tests your understanding of both Ohm's Law and RMS-to-peak conversion.
  • Solve problems involving simple resistive circuits, then check your work against the common mistake of thinking resistance changes with frequency.
4 🌍 Connect to Real-World Physics
  • Examine the APPLICATIONS section. Explain to a friend how a space heater uses a resistive element to convert AC electrical energy into heat.
  • Look at an incandescent light bulb, as mentioned in the APPLICATIONS. Identify the filament as the resistor and describe its role in the circuit.
  • Find the power rating on a real-world appliance like a toaster. Use the standard wall voltage (120V RMS) to calculate its effective resistance.
  • Consider the Real-World Examples. Discuss why the simple resistive model is a good approximation for devices like electric kettles but not for electric motors.
Master Ohm's Law for resistive AC circuits by understanding its in-phase nature, practicing with RMS values, and seeing it power everyday devices.

Frequently Asked Questions

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