# AC Voltage Resistor

## 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 current that varies with time periodically, then this current is called alternating current. Here, we are considering that the source is producing sinusoidal varying potential difference or voltage across its terminals. We can represent it as

$V = V_{m}sin \omega t$

Where $V_{m}$ 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.

The above image represents the circuit diagram of a resistor connected to a source which is producing AC voltage.

### AC Voltage Applied to a Resistor Equation

To calculate the current through the resistor due to the present voltage source, we will apply the Kirchhoff's loop rule which is

$\sum V(t) = 0$

By using this equation, we can write

$V = V_{m}sin \omega t = iR$

Or,

$i = \frac{V_{m}}{R} sin \omega t$

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 $i = \frac{V_{m}}{R} sin \omega t$ as

$i = i_{m}sin \omega t$

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 are in phase with each other. Also, we know that $p = i^{2}R$. Hence, we can write

$p = i_{m}^{2} R \omega t$

We can represent the average value of power generated in the complete cycle as

$\overline{p} = i^{2} R = i_{m}^{2} R \omega t$

The above quantities $i_{m}$ and R are constant.

Using trigonometry, $\omega t = \frac{1}{2} (1 - cos 2 \omega t)$

From the above graph, we can see that cos2ꞷt = 0. Now, we put this value of cos2ꞷt 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 $\overline{p} = i_{m}^{2} R \omega t$. 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 doesn't mean that there is no dissipation of electrical energy. We know that the joule heating is $i^{2}R$, and thus it is always positive because it depends on$i^{2}$. 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

$W_{2A} = (2A)^{2} R (1s)$

$W_{2A} = 4R$

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,

$I_{0} = (4A)\times \sqrt{2}$

$I_{0} = 5.6 A$