Alternating Current (AC) Through Resistors: Formula, Power, RMS & Examples

23.4 ALTERNATING CURRENTS THROUGH RESISTOR

23.4.1 Ohm’s Law for Resistive Circuit

Figure 23.6 The figure shows a resistor connected to an alternating voltage source, where the applied voltage produces an alternating current that follows Ohm’s law.

Suppose an alternating voltage V is applied across a resistor of resistance R, as shown in Figure 23.6. Then the voltage causes an alternating current I to flow through the resistor. The voltage and current obeys Ohm’s law, so that we have

$V = IR$

23.4.2 The Rotating Phasor

Figure 23.7 The figure shows a rotating arrow (phasor) with fixed length about the origin, representing a sinusoidal current or voltage with constant angular velocity.

1.  Other than using the sine or cosine graph to deduce the instantaneous values and phase of a sinusoidal current or voltage, we can also use an alternative method, that is, the phasor method.

2.  Consider an arrow of fixed length, with its tail placed at the origin O of the Cartesian coordinate system, as shown in Figure 23.7. It rotates about O at a certain constant angular velocity. This rotating arrow is referred to as the phasor.

3.  A phasor is used to represent a sinusoidal current and voltage.

4.  In order to use the rotating arrow to represent a sinusoidal current I or voltage V, we must have the following:

(a) The length OA of the arrow is used to represent the maximum value of the current ($I_0$) or voltage ($V_0$).

(b) The arrow rotates in the anti-clockwise direction with an angular velocity $ω$ which is the same as that of the current or voltage.

(c) The vertical projection (i.e. the component length) OB of the arrow on the YY′-axis represents the instantaneous value of I or V.

(d) The angle $θ$ between the XX′-axis and the arrow represents the phase angle of I or V.

EXAMPLE 23.3

Refer to Figure 23.7. The arrow OA is used to represent a sinusoidal current. A scale of 1 cm : 1 A is used. The current I is given by the following expression:

$I = (2.0 \ A) \ sin \ (100πt)$

1. What is the length of the arrow?
2. How fast is the arrow rotating?
3. At a particular instant, the angle θ is 50°. What is the current at that instant? Is the current increasing or decreasing after that instant?

Answer

(a) The maximum value of the current is 2.0 A. Hence, the length of the arrow is 2.0 cm.

(b) The current has angular velocity $ω = 100π$.
Hence, the arrow rotates in the anti-clockwise direction with an angular velocity of $100π \ rad s^{-1}$.

(c) The projected length OB on the YY′-axis is

$OB = OA \ sin \ θ$

$= (2 cm) \ sin \ 50^0 = 1.5 \ cm$

The instantaneous current is 1.5 A.

Since the arrow rotates in the anti-clockwise direction, the projected length OB increases as the angle θ increases. Hence, the current increases.

23.4.3 The Phasor Incorporated in Sine/Cosine Graph

Figure 23.8 The figure shows a rotating phasor linked to a sine or cosine graph, illustrating how its rotation represents a sinusoidal current or voltage over time.

The rotating phasor can be incorporated in the sine or cosine graph which is used to represent a sinusoidal current or voltage, as shown in Figure 23.8.

Notice that:

1. the projected length I on the YY′-axis represents the instantaneous current/voltage at time t₁.
2. The arrow points vertically upwards or downwards at the instant when the current/voltage is a maximum.
3. The arrow completes one round when the current/voltage has undergone one cycle of change.

23.4.4 Phase Difference between I and V for Pure Resistors

Figure 23.9 The figure shows a resistor connected to a sinusoidal voltage source, where the applied voltage produces an alternating current through the resistor.

A sinusoidal voltage V is applied to a resistor which has resistance R, as shown in Figure 23.9. The voltage produces an alternating current I through the resistor. Let the voltage be represented by the expression

$V = V_0 \ sin \ ωt$

where V₀ is the maximum value of the voltage. Since the circuit is a resistive one, we have

$V = IR$

or,

$I = \frac{(V_0 \ sin \ ωt)}{R}$

$=\left(\frac{V_0}{R}\right) \ sin \ ωt$

$= I_0 \ sin \ ωt$

where the maximum current

$I_0 = \frac{V_0}{R}$

or,

$V_0 = I_0R$

We notice that V and I both have one same term, i.e. sin ωt. This means that for a resistive circuit, the voltage and current are in phase with each other, or φ = 0°, where φ is the phase difference between V and I.

Figure 23.10 The figure shows two overlapping phasors representing voltage (V) and current (I), indicating that they are in phase and vary together.

We can illustrate this phase relationship between V and I using the phasor diagram. Refer to Figure 23.10. We notice that:

1. there are two arrows, one representing V and the other representing I.
2. the two arrows overlap each other and are always together in that way, indicating that V and I are in phase.
3. The arrows rotate with the same angular velocity.
4. I = 0 when V = 0, and I is maximum when V is maximum. In other words, I varies in step with V. This is the main characteristic of two quantities varying in phase or with a phase difference of zero.
5. I is negative (or positive) when V is negative (or positive). This means that the product IV is always positive for a pure resistor. This is not true for a capacitor or an inductor.

23.4.5 Power Supplied to Resistor by AC Voltage

Suppose an alternating voltage V is applied across a resistor of resistance R. The electrical power P delivered to the resistor is given by

$P = IV$

Since the circuit is a resistive circuit, we have

$V = IR$

Hence, we may express the equation for P as

$P = I^2R$

or,

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

23.4.6 Power Supplied to Resistor by Sinusoidal Voltage Supply

1. Instantaneous power received by resistor

Figure 2311 The figure shows the variation of instantaneous power with time, where the power remains positive and has twice the frequency of the current, indicating continuous energy dissipation as heat.

Suppose a sinusoidal voltage V is applied across a resistor of resistance R where

$V = V_0 \ sin \ ωt$

A sinusoidal current I passes through the resistor. Since V and I are in phase, the current is given by

$I = I_0 \ sin \ ωt$

The instantaneous electrical power P delivered to the resistor is given by

$P = IV$

$= (I_0 \ sin \ ωt)(V_0 \ sin \ ωt)$

$P = (I_0V_0) \ sin^2 \ (ωt)$

Figure 23.11 shows how the instantaneous power P varies with time t. Notice the following:

1. The curve for P does not go below the negative region. In other words, the power received by the resistor is always positive. This means that the power received is totally dissipated as heat. None of the power received is returned to the circuit.
2. When the current has undergone one cycle, the power curve goes through two peaks.

2. Average power received by resistor

We can show that

average value of $sin^2 \ (ωt) = \frac{1}{2}$

Hence, the average power $P_{av}$ received by the resistor is given by

$P_{av} = \frac{1}{2} I_0V_0$

Since we have

$V_0 = I_0R$

we may write the equation above as

$P_{av} = \frac{1}{2} I_0^2R$

or,

$P_{av} = \frac{1}{2} \frac{V_0^2}{R}$

23.4.7 Quantity of Heat Dissipated by Resistor

The electrical energy received by the resistor is converted to heat. Hence, the power dissipated by the resistor in the form of heat is

$P = IV$

$= \frac{dQ}{dt}$

where dQ/dt is the rate of production of heat.

The quantity of heat Q produced in the time interval between t₁ and t₂ is given by

$Q = \int_{t_1}^{t_2}P dt$

 = area below the curve in a P–t graph

Figure 23.12 shows an example of a P–t graph. The shaded area between t₁ and t₂ represents the amount of heat produced in that time interval.

Figure 23.12 The figure shows a power-time graph where the shaded area between two time intervals represents the amount of heat energy produced.

EXAMPLE 23.4

An alternating voltage V with waveform as shown in Figure 27.13 is applied across a 10 Ω resistor. Determine

1. the electrical power received by the resistor at time t = 0.20 ms
2. the amount of heat dissipated by the resistor in the time interval between t = 0 and t = 1.0 ms.

Figure 27.13 The figure shows a sinusoidal voltage applied across a resistor, used to determine the instantaneous electrical power at a specific time.

Answer

(a) gradient $= \frac{10 \ V}{1 \ ms} = (V + 10)/t$

At time t = 0.20 ms,      V = −8.0 V

The electrical power P received by the resistor at time t = 0.20 ms is given by

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

$= \frac{8^2}{10}= 6.4 \ W$

(b) Between t₁ = 0 and t₂ = 1.0 ms, the voltage V can be expressed as

$V = 10t − 10 = 10(t − 1) V$

The instantaneous power P is given by

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

$= \frac{10^2(t − 1)^2}{10}$

$= 10 (t^2 − 2t + 1)$

The amount of heat Q produced by the resistor in the time interval between t₁ = 0 and t₂ = 1.0 ms is given by

$Q = \int_{t_1}^{t_2}P dt$

t₂ = 1 ms

$Q = \int_{t_1=0}^{t_2}(t^2-2t+1) dt$

$= 10 \left[\frac{t^3}{3} − 2\frac{t^2}{2}+ t \right]_0^{1 \ ms}$

$= 10 (1 \times 10^{-3}) \ J = 10 \ mJ$

Figure 23.14 shows how the electrical power varies with time as the voltage changes with time. The shaded area between 0 and 1 ms represents the amount of heat produced in that time interval.

Figure 23.14 The figure shows how electrical power varies with time, where the shaded area represents the amount of heat produced over the given time interval.

shaded area = area below the average power line and between 0 and 1 ms

The average power is a constant. Hence, we may imagine that this average power is generated by an equivalent constant current (dc) that would flow through the same resistor. This concept is used in defining the rms value of an ac.

EXAMPLE 23.5

An alternating voltage V is given by

$V = (12 \ V) \ sin \ (120πt)$

It is applied across a 10 Ω resistor. Determine

1. the average electrical power supplied to the resistor
2. the amount of heat produced by the resistor in 2 minutes.

Answer

(a) Since the voltage is a sinusoidal voltage, we have

$P_{av} = \frac{1}{2} \frac{V_0^2}{R}$

where V₀ = 12 V and R = 10 Ω

$P_{av}= \frac{1}{2}\frac{12^2}{10} = 7.2 \ W$

(b)

$P_{av} =$ heat produced / time taken

$=\frac{Q}{t}$

$Q = P_{av}t$

$= (7.2 Js^{-1})(2 \times 60 \ s) = 864 \ J$

EXAMPLE 23.6

A current I which is given by

$I = (20 A) \ sin \ 100πt$

flows through a 20 Ω resistor. Determine

1. the maximum voltage that can be applied across the resistor
2. the amount of heat produced by the resistor in 10 minutes

Answer

(a) Since the current is a sinusoidal current, we have

$V_0 = I_0R$

$= (20)(20) = 400 \ V$

(b) The amount of heat Q produced by the resistor is given by

$Q = P_{av}t$

$= \frac{1}{2} I_0^2R)t$

$= \frac{1}{2} (20^2)(20)(10 \times 60) = 2.4 \times 10^6 \ J$

EXAMPLE 23.7

An electric heater with resistance of 10 Ω is connected to a sinusoidal voltage which has a maximum voltage of 240 V. The heater is totally immersed in 3.0 kg of water at 20 °C. Determine the temperature of the water 5 minutes after switching on the heater. State the assumptions made in the calculation. (Specific heat capacity of water = 4.20 kJ kg⁻¹ K⁻¹)

Answer

In time t = 5 minutes, the amount of heat Q produced by the heater is given by

$Q = P_{av}t$

where P_av is the average electrical power supplied to the heater. Since the voltage is sinusoidal, we have

$P_{av} = \frac{1}{2}\frac{V_0^2}{R}$

$= \frac{1}{2}\frac{240^2}{10} = 2.88 \times 10^2 \ J \ s^{-1}$

Hence,

$Q = (2.88 \times 10^3)(5 \times 60) = 8.64 \times 10^5 \ J$

We have

$Q = mc (θ_1 − θ_0)$

$θ_1 − θ_0 = \frac{Q}{mc}$

$= \frac{8.64 \times 10^5}{(3.0)(4.20 \times 10^3)} = 68.6 \ ^0C$

$θ_1 = 68.6 + θ_0$

$= 68.6 + 20 = 88.6 \ ^0C$

Assumptions:
(a) The electrical energy supplied to the resistor is completely converted to heat.
(b) The heat produced is completely absorbed by the water.

EXAMPLE 23.8

A resistor of resistance 20 Ω produces 2.50 kJ of heat at a constant rate in 4 minutes when a sinusoidal voltage is applied across it. Determine

1. the maximum voltage that can be applied across the resistor
2. the maximum current that can flow through the resistor.

Answer

(a) The average power $P_{av}$ of heat production is given by

$P_{av}$ = heat produced/time taken

$= \frac{2500}{(4 \times 60)} = 10.4 \ J \ s^{-1}$

Since the applied voltage is a sinusoidal voltage, we have

$P_{av} = \frac{1}{2} \frac{V_0^2}{R}$

$10.4 = \frac{V_0^2}{2R}$

$V_0 = 20.4 \ V$

(b) Since the applied voltage is a sinusoidal voltage, we have

$V_0 = I_0R$

$20.4 = I_0(20)$

$I_0 = 1.02 \ A$

EXAMPLE 23.9

A current I given by the equation

$I = (4.0 \ A) \ cos (314t)$

passes through a resistor. The resistor produces 1.5 kJ of heat in 3 minutes. Determine the resistance of the resistor.

Answer

The average power $P_{av}$ of heat production by the resistor is

$P_{av} =$ heat produced/time taken

$= \frac{1500}{3 \times 60} = 8.33 \ J \ s^{-1}$

Since the current is a sinusoidal current, we have

$P_{av} = \frac{1}{2} I_0^2R$

$8.33 = \frac{1}{2} (4.0)^2R$

$R = 1.04 \ Ω$

23.4.8 The RMS Value of Alternating Current

Figure 23.15 The figure shows an alternating current replaced by an equivalent direct current that produces the same heating effect, defining the rms value of the current.

1. Suppose an alternating current flows through a resistor of resistance R. Then a certain amount of heat is produced every second by the resistor. Let the amount of heat generated per second, i.e. the power, be P.
2. We can replace the alternating current with a direct current I which also produces heat as it flows through the same resistor. The value of the direct current can be adjusted until the heat produced per second by the resistor is also P, as shown in Figure 23.15. In other words, we obtain a value of the dc which produces the same amount of heat per second as that produced by the ac. This value of the dc, I_rms, is known as the root mean square (rms) value of the alternating current. We have

power produced by $I_{rms}$ = power produced by ac

for a given resistance.

3. The rms value of an alternating current is the equivalent direct current which produces the same power dissipation in a given resistance.

EXAMPLE 23.10

Figure 23.16 The figure shows the waveform of an alternating current used to determine its rms value based on its variation over time.

Determine the rms value of the alternating current whose waveform is as shown in Figure 23.16.

Answer

Figure 23.17 The figure shows the graph of current squared against time, illustrating the variation over one period of 2.0 s for rms calculation.

Square the current value.
Figure 23.17 shows how the I²–t graph looks like. The period is T = 2.0 s.

power dissipated by $I_{rms}$ = power dissipated by ac

heat produced in period T by $\frac{I_{rms}}{T}$=heat produced in period T by ac/T

heat produced in period T by $I_{rms}$ = heat produced in period T by ac

$(I_{rms}^2 R)T = (I_0^2R)T$

where $I_0 = 2.0 A$

$I_{rms}^2= I_0^2$

$I_{rms}= I_0$

$= 2.0 A$

EXAMPLE 23.11

Determine the rms value of the alternating current whose waveform is as shown in Figure 23.18.

Figure 23.18 The figure shows the waveform of an alternating current used to determine its rms value over one complete cycle.

Answer

Square the current value.

Figure 23.19 The figure shows the graph of current squared against time over a period of 2.0 s, used to determine the rms value.

Figure 23.19 shows how the I²–t graph looks like. The period is T = 2.0 s.

power dissipated by $I_{rms}$ = power dissipated by ac

heat produced in period T by $\frac{I_{rms}}{T}$= heat produced in period T by ac/T

heat produced in period T by $I_{rms}$ = heat produced by ac between 0 and $\frac{1}{2}T$ + heat produced by ac between $\frac{1}{2}T$ and T

$(I_{rms}R^2)T = (I_1^2R)\left(\frac{1}{2} T\right) + (I_2^2R)\left(\frac{1}{2}T\right)$

where $I_0 = 2.0 \ A$ and $I_0 = −1.0 \ A$.

$I_{rms}^2 \times T = I_1^2 \times \left(\frac{1}{2}T \right) + I_2^2 \times \left(\frac{1}{2}T\right)$

Figure 23.20 The figure shows areas under current-squared graphs, where areas A and B represent alternating current segments and area C represents the equivalent rms current over one period.

Refer to Figure 23.20. We have

area $C = I_{rms}^2 \times T$

area $A = I_1^2 \times \left(\frac{1}{2}T \right)$

area $B = I_2^2 \times \left(\frac{1}{2}T\right)$

area C = area A + area B

$(I_{rms}^2)(2 \ s) = (4.0 \ A^2)(1 \ s) + (1.0 \ A^2)(1 \ s) = 5.0 A^2 \ s$

$I_{rms}^2 = 2.5 \ A^2$

$= \sqrt{2.5} = 1.58 \ A$

23.4.9 The RMS Value of Sinusoidal Current and Voltage

1. RMS current

A sinusoidal voltage V given by the equation

$V = V_0 \ sin \ ωt$

is applied across a resistor of resistance R and produces a current I which flows through the resistor. The current is given by

$I = I_0 \ sin \ ωt$

Then the rate of production of heat $P_{av}$ by the resistor is

$P_{av} = \frac{1}{2} I_0V_0$

$P_{av} = \frac{1}{2}I_0^2R$      ($V_0=I_0R$)

Suppose that we replace the sinusoidal voltage with another voltage which produces a constant current. Let the magnitude of this constant current flowing through the same resistor be such that the resistor also produces heat at the same rate, $P_{av}$. Then, by definition, this constant current $I_{rms}$ must be the rms value of the sinusoidal current. We now have

$P_{av} = I_{rms}^2R$

$I_{rms}^2R = \frac{1}{2} I_0^2R$

$I_{rms} = \frac{I_0}{\sqrt{2}}$

2. RMS voltage

The rms value of the alternating voltage is

$V_{rms} = \frac{V_0}{\sqrt{2}}$

3. Average power in terms of rms current and voltage

We have

(a)   $P_{av} = I_{rms}^2R$

(b)   $P_{av} = \frac{1}{2}I_0V_0$

$= \left(\frac{1}{\sqrt{2}}\right)^2I_0V_0$

$= \left(\frac{I_0}{\sqrt{2}}\right)\left(\frac{V_0}{\sqrt{2}}\right)$

$P_{av} = I_{rms}V_{rms}$

EXAMPLE 23.12

A current I represented by the expression

$I = (5.0 \ A) cos \ 314t$

flows through a resistor of resistance 5.0 Ω. Determine

1. the maximum current which can flow through the resistor
2. the rms value of the alternating current
3. the amount of heat produced by the resistor in 1.5 minutes.

Answer

(a) Maximum current

$I_0 = 5.0 \ A$

(b) rms value

$I_{rms}= \frac{I_0}{\sqrt{2}}$

$= \frac{5.0}{\sqrt{2}}= 3.54 \ A$

(c) Amount of heat produced

$Q = P_{av}t$

$= (I_{rms}^2R)t$

$= (3.54^2)(5.0)(1.5 \times 60) = 5.6 \times 10^2 \ J$

EXAMPLE 23.13

An electrical appliance draws 1.50 kW of power from the 240 V mains. Determine the current that passes through the appliance. State any assumptions made.

Answer

1. The voltage of ‘240 V mains’ is assumed to be the rms voltage.
2. The power of 1.50 kW received by the appliance is assumed to be the average power.
3. The current is assumed to pass through a pure resistor in the appliance.

$P_{av} = I_{rms} V_{rms}$

$1500 = I_{rms}(240)$

$I_{rms} = 6.3 \ A$

EXAMPLE 23.14

A sinusoidal current with a maximum value of 10 A flows through a 5 Ω resistor. Determine the average electrical power delivered to the resistor.

Answer

$P_{av}= I_{rms}^2R$

$= \left(\frac{I_0}{\sqrt{2}}\right)^2R$

$= \left(\frac{10}{\sqrt{2}}\right)^2(5) = 250 \ W$

EXAMPLE 23.15

A coil of area 2.0 cm² has 20 turns. It is placed in a uniform magnetic field of flux density 50 mT. It then rotates about an axis which passes through its centre and perpendicular to the field at frequency 5 revolutions per second. Determine

1. the maximum voltage produced across the coil
2. the rms value of the voltage across the coil.

Answer

(a) Maximum voltage

$V_0 = BANω$

$= (50 \times 10^{-3})(2.0 \times 10^{-4})(20)(5 \times 2π)$

$= 6.28 \ mV$

(b) rms voltage

$V_{rms} = \frac{V_0}{\sqrt{2}}$

$= \frac{6.28}{\sqrt{2}}= 4.44 \ mV$

EXAMPLE 23.16

A sinusoidal voltage supply is connected to a resistor. A voltmeter records a reading of 5.5 V across the resistor. The deflection of an ammeter pointer indicates that there is a current of 2.3 A flowing through the resistor. Both instruments record rms values. Determine

1. the maximum voltage that can be applied across the resistor
2. the maximum current that can flow through the resistor
3. the average electrical power supplied to the resistor
4. the resistance of the resistor

Answer

(a) $V_0 = \sqrt{2} V_{rms}$

$= \sqrt{2} (5.5) = 7.78 \ V$

(b) $I_0 = \sqrt{2} I_{rms}$

$= \sqrt{2}(2.3) = 3.25 \ A$

(c) $P_{av} = I_{rms}V_{rms}$

$= (2.3)(5.5) = 12.7 \ W$

(d) $P_{av} = I_{rms}^2R$

$12.7 = (2.3)^2R$

$R = 2.4 \ Ω$

⚡ Applications of Alternating Current (AC) in Daily Life

💡

Household Electricity

Alternating current is used in homes to power appliances such as lamps, fans, televisions, and refrigerators because it can be transmitted efficiently over long distances.

🔌

Power Transmission

AC is used in electrical grids because its voltage can be easily increased or decreased using transformers, reducing energy loss during transmission.

🔥

Electric Heating

Devices like electric heaters, irons, and kettles use AC where electrical energy is converted into heat based on power formula P = I²R.

⚙️

Electric Motors

AC motors are widely used in industrial machines, fans, and washing machines due to their efficiency and simple construction.

📻

Communication Systems

Alternating signals are used in radio, television, and wireless communication to transmit information through electromagnetic waves.

🔋

RMS Value Usage

The RMS value helps determine the equivalent DC power in AC circuits, ensuring safe and effective design of electrical devices.

⚡ Conclusion of Alternating Current (AC)

Alternating current (AC) is a fundamental concept in electricity where the current changes direction periodically. It plays a vital role in modern electrical systems due to its efficiency in power transmission and its wide range of practical applications.

• AC can be easily transmitted over long distances with minimal energy loss.
• The concept of RMS value allows AC to be compared with DC in terms of power.
• In resistive circuits, voltage and current are in phase.
• Electrical power in AC circuits depends on current, voltage, and resistance.
• AC is widely used in household appliances, industries, and communication systems.

Key Insight: The RMS value of AC enables us to calculate power effectively, making AC the most practical and widely used form of electrical energy in everyday life.

❓ Frequently Asked Questions (FAQ) - Alternating Current

What is alternating current (AC)?

Alternating current (AC) is an electric current that periodically reverses direction and changes its magnitude over time.

What is the difference between AC and DC?

AC changes direction periodically, while direct current (DC) flows in only one direction with constant magnitude.

What is RMS value in AC?

The RMS (root mean square) value is the equivalent DC value that produces the same power in a resistor as the AC current.

Why is AC used for power transmission?

AC is used because its voltage can be easily transformed to high values, reducing energy loss during long-distance transmission.

What is the formula for RMS current?

The RMS current is given by I_rms = I₀ / √2 for a sinusoidal current.

Are voltage and current in phase in resistive circuits?

Yes, in a pure resistive circuit, voltage and current are in phase, meaning they reach their maximum and minimum values at the same time.

What is average power in AC circuits?

The average power is given by P = I_rms V_rms or P = I_rms²R.

What is frequency in AC?

Frequency is the number of cycles completed per second, measured in hertz (Hz).

What is a sinusoidal current?

A sinusoidal current is an alternating current that varies with time following a sine wave pattern.

Where is AC used in daily life?

AC is used in household electricity, industrial machines, communication systems, and many electrical devices.

Share :