We also saw that sinusoidal waveforms and functions that were previously drawn in the *time-domain* transform can be converted into the spatial or *phasor-domain* so that phasor diagrams can be constructed to find this phasor voltage-current relationship.

Now that we know how to represent a voltage or current as a phasor we can look at this relationship when applied to basic passive circuit elements such as an **AC Resistance** when connected to a single phase AC supply.

Any ideal basic circuit element such as a resistor can be described mathematically in terms of its voltage and current, and in our tutorial about Resistors , we saw that the voltage across a pure ohmic resistor is linearly proportional to the current flowing through it as defined by Ohm’s Law . Consider the circuit below.

### AC Resistance with a Sinusoidal Supply

When the switch is closed, an AC voltage, V will be applied to resistor, R. This voltage will cause a current to flow which in turn will rise and fall as the applied voltage rises and falls sinusoidally. As the load is a resistance, the current and voltage will both reach their maximum or peak values and fall through zero at exactly the same time, i.e. they rise and fall simultaneously and are therefore said to be “*in-phase* ”.

Then the electrical current that flows through an AC resistance varies sinusoidally with time and is represented by the expression, I(t) = Im x sin(ωt + θ), where Im is the maximum amplitude of the current and θ is its phase angle. In addition we can also say that for any given current, i flowing through the resistor the maximum or peak voltage across the terminals of R will be given by Ohm’s Law as:

and the instantaneous value of the current, i will be:

So for a purely resistive circuit the alternating current flowing through the resistor varies in proportion to the applied voltage across it following the same sinusoidal pattern. As the supply frequency is common to both the voltage and current, their phasors will also be common resulting in the current being “in-phase” with the voltage, ( θ = 0 ).

In other words, there is no phase difference between the current and the voltage when using an AC resistance as the current will achieve its maximum, minimum and zero values whenever the voltage reaches its maximum, minimum and zero values as shown below.

### Sinusoidal Waveforms for AC Resistance

This “in-phase” effect can also be represented by a phasor diagram. In the complex domain, resistance is a real number only meaning that there is no “j” or imaginary component. Therefore, as the voltage and current are both in-phase with each other, there will be no phase difference ( θ = 0 ) between them, so the vectors of each quantity are drawn super-imposed upon one another along the same reference axis. The transformation from the sinusoidal time-domain into the phasor-domain is given as.

### Phasor Diagram for AC Resistance

As a phasor represents the RMS values of the voltage and current quantities unlike a vector which represents the peak or maximum values, dividing the peak value of the time-domain expressions above by √2 the corresponding voltage-current phasor relationship is given as.

### RMS Relationship

### Phase Relationship

This shows that a pure resistance within an AC circuit produces a relationship between its voltage and current phasors in exactly the same way as it would relate the same resistors voltage and current relationship within a DC circuit. However, in a DC circuit this relationship is commonly called **Resistance**, as defined by Ohm’s Law but in a sinusoidal AC circuit this voltage-current relationship is now called **Impedance**. In other words, in an AC circuit electrical resistance is called “Impedance”.

In both cases this voltage-current ( V-I ) relationship is always linear in a pure resistance. So when using resistors in AC circuits the term **Impedance**, symbol **Z** is the generally used to mean its resistance. Therefore, we can correctly say that for a resistor, DC resistance = AC impedance , or R = Z.

The impedance vector is represented by the letter, ( Z ) for an AC resistance value with the units of Ohm’s ( Ω ) the same as for DC. Then Impedance ( or AC resistance ) can be defined as:

### AC Impedance

Impedance can also be represented by a complex number as it depends upon the frequency of the circuit, ω when reactive components are present. But in the case of a purely resistive circuit this reactive component will always be zero and the general expression for impedance in a purely resistive circuit given as a complex number will be:

Z = R + j0 = R Ω’s

Since the phase angle between the voltage and current in a purely resistive AC circuit is zero, the power factor must also be zero and is given as: cos 0^{o} = 1.0 , Then the instantaneous power consumed in the resistor is given by:

However, as the average power in a resistive or reactive circuit depends upon the phase angle and in a purely resistive circuit this is equal to θ = 0, the power factor is equal to one so the average power consumed by an AC resistance can be defined simply by using Ohm’s Law as:

which are the same Ohm’s Law equations as for DC circuits. Then the effective power consumed by an AC resistance is equal to the power consumed by the same resistor in a DC circuit.

Many AC circuits such as heating elements and lamps consist of a pure ohmic resistance only and have negligible values of inductance or capacitance containing on impedance.

In such circuits we can use both Ohm’s Law ,Kirchoff’s Law as well as simple circuit rules for calculating and finding the voltage, current, impedance and power as in DC circuit analysis. When working with such rules it is usual to use RMS values only.

## AC Resistance Example No1

An electrical heating element which has an AC resistance of 60 Ohms is connected across a 240V AC single phase supply. Calculate the current drawn from the supply and the power consumed by the heating element. Also draw the corresponding phasor diagram showing the phase relationship between the current and voltage.

1. The supply current:

2. The Active power consumed by the AC resistance is calculated as:

3. As there is no phase difference in a resistive component, ( θ = 0 ), the corresponding phasor diagram is given as:

## AC Resistance Example No2

A sinusoidal voltage supply defined as: V(t) = 100 x cos(ωt + 30^{o}) is connected to a pure resistance of 50 Ohms. Determine its impedance and the peak value of the current flowing through the circuit. Draw the corresponding phasor diagram.

The sinusoidal voltage across the resistance will be the same as for the supply in a purely resistive circuit. Converting this voltage from the time-domain expression into the phasor-domain expression gives us:

Applying Ohms Law gives us:

The corresponding phasor diagram will therefore be:

## Impedance Summary

In a pure ohmic **AC Resistance**, the current and voltage are both “in-phase” as there is no phase difference between them. The current flowing through the resistance is directly proportional to the voltage across it with this linear relationship in an AC circuit being called **Impedance**.

Impedance, which is given the letter Z, in a pure ohmic resistance is a complex number consisting only of a real part being the actual AC resistance value, ( R ) and a zero imaginary part, ( j0 ). Because of this Ohm’s Law can be used in circuits containing an AC resistance to calculate these voltages and currents.

In the next tutorial about AC Inductance we will look at the voltage-current relationship of an inductor when a steady state sinusoidal AC waveform is applied to it along with its phasor diagram representation for both pure and non-pure inductance’s.

What are the effect resistance has on current in an AC circuit

The same as it does in a DC circuit. Higher resistance less current.

I have problem. How to convert source impedance into a pure resistance to solve maximum power transfer example?

If its AC, make XL = XC at the given frequency for series resonance, leaving just R

If its DC, then Z^2 = R^2 + (XL – XC)^2 and make Z equal the load resistance, RL

Hi, I have a question.in the second example Shouldn’t we find the Vrms of 100 first (100/√2=70,7 V) and then calculate Irms from Vrms for the resistance(70,7/50=1,414 A)? Thanks

Hello Lefteris, Yes you could if you so wished as this gives the rms value, the same as 2amps(peak) x (1/√2) = 1.414 amps (rms).