AC Voltage Applied to a Resistor
Resistance is the opposition to the flow of the current offered by any substance. The substance that limits the flow of electric current in the circuit is the resistor. If the source is producing a current that varies with time periodically, then this current is called an alternating current. Here, we are considering that the source is producing sinusoidal varying potential differences or voltage across its terminals. We can represent it as
V = Vm Sin ωt
where Vm represents the amplitude of the oscillating potential difference and represents the angular frequency. In the below article, we will describe what will happen when we apply the AC voltage source to a resistor.
(image will be uploaded soon)
The above image represents the circuit diagram of a resistor connected to a source that is producing AC voltage.
AC Voltage Applied to a Resistor Equation
Please note that the resistance value of the resistor in both AC and DC circuits is likewise without considering the frequency of the AC supply voltage. Also, the alteration in direction of current in AC supply does not impact resistors' behaviour.
To calculate the current through the resistor due to the present voltage source, we will apply Kirchhoff's loop rule which is
∑ V(t)=0
By using this equation, we can write
V = Vm Sin ωt = iR
Or,
i = Vm RSin ωt
where i represents the current, and R represents the resistance of the given resistor.
According to ohm's law,
V = iR
Or
\[i = \frac{V}{R}\]
Ohm law works equally where the source is producing AC Voltage or DC voltage. Hence, we can also write the above equation as
\[i = \frac{V_{m}}{R}\]
By using this equation, we can also write \[i = \frac{V_{m}sin \omega t}{R} \] as;
i = im nSi ωt
(image will be uploaded soon)
The above image shows the graphical representation of the variation in current and voltage with respect to time.
From the above image, it is clear the current and voltage reach maximum and minimum values at the same time. Both the current passing through the resistor and voltage across it are the sinusoidal quantities and are in phase with each other.
Average Value of the Power
AC Voltage Applied to a Resistor derivation shows us the potential difference, and current is in phase with each other. Also, we know that p = i2R. Hence, we can write:
p = im2Rωt
We can represent the average value of power generated in the complete cycle as;
p = i2R = im2Rωt
The above quantities im and R are constant.
Using trigonometry, \[\omega t = \frac{1}{2} (1 - cos 2 \omega t)\]
From the above graph, we can see that cos2ωt = 0. Now, we put this value of cos2ωt in the above equation. Now, we get:
\[\omega t = \frac{1}{2} (1 - 0)\]
Or,
\[\omega t = \frac{1}{2}\]
Now, we will put this value of ωt in the equation p = im2Rωt. Thus we get:
\[\overline{p} = \frac{1}{2} i_{m}^{2} R \omega t\]
It is also important to note we can express AC power into DC power by denoting the current in terms of root-mean = square current or effective current.
\[I_{rms} = \sqrt{i^{-2}} = \sqrt{\frac{1}{2} i_{m}^{2}}\]
By evaluating, we can also write the above equation as;
\[I_{rms} = \frac{i_{m}}{\sqrt{2}} = 0.707 i_{m}\]
Note on Power Consumption in AC Voltage Applied to a Resistor
The sum of instantaneous current over one complete cycle is zero, and hence, the average current is also zero. However, it does not mean that there is no dissipation of electrical energy. We know that the joule heating is i2R, and thus it is always positive because it depends on i2. The thermal energy developed during the time t to t+dt in the resistor is:
\[i^{2}Rdt = i_{m}^{2} (\omega t) Rdt\]
The thermal energy developed in one time period is:
U = \[\int i^{2} Rdt = \int i_{m}^{2} (\omega t) Rdt\]
Over the limit 0 to T
= R\[\int i_{m}^{2} (\omega t) Rdt\]
= \[\int i_{rms}^{2} (\omega t) Rdt\]
It means that the root-mean-square value of alternating current is the steady current that would generate the same amount of heat in the resistor in a given amount.
Solved Examples
Example 1: What is the approx peak value of an alternating current producing four times the heat produced per second by a steady current of 2 A in a resistor?
Ans. The equation that represents the relation between RMS current and peak current is:
\[I_{0} = I_{rms} \sqrt{2}\]
The expression for the heat produced by the resistor is
W = \[I^{2} Rt\]
Hence, the heat generated by the resistor when the steady current of 2A flows through it is
\[W_{2A} = (2A)^{2} RT\]
\[W_{2A} = 4R\]
The heat produced by the alternating current is four times the heat generated by the constant current. Hence,
\[W_{alternating} = 4\times W_{2A}\]
\[W_{alternating} = (4\times 4) R = 4^{2}R\]
Therefore, the RMS value of the alternating current is 4A.
Hence,
\[I_{0} = (4A)\times \sqrt{2}\]
\[I_{0} = 5.6 A\]
FAQs on AC Voltage Resistor
1. Why is the average value of alternating current over the complete cycle zero?
The average value of the current is the total of all the instantaneous values of potential difference or current over the complete cycle. In the DC system, the potential difference and current remain constant over a period of time. Hence, one can quickly determine its magnitude. However, the alternating current varies from time to time. In alternating current, the magnitude and the direction of the current changes after every half cycle. Since the positive current flows in the positive half cycle and negative current flows in the other half cycle. Therefore, their sum is zero, as both are symmetrical, and the two halves cancel each other.
2. What is the difference between RMS voltage and average voltage?
The mean value of a periodic wave is the average of all the instantaneous areas taken under the curve over a given period of the waveform. In the case of sinusoidal quantities like alternating current, we take the period as one-half of the wave cycle. The root means square (RMS) value is the square root of the mean of the squares of the instantaneous values taken over a complete cycle. For a pure sinusoidal waveform, we can calculate the RMS voltage as 0.707 x peak value. Moreover, the average value of the alternating current over a complete cycle is 0.
