Question

For a given ac, $i = {i_m}\sin \omega t$ show that average power dissipated is $\dfrac{1}{2}{i^2}R$.

Answer 1

Alternating current is said to be better than direct current because of its ability to be converted into higher or lower values and hence used in transformers to travel long distances unlike dc. The power dissipated for an alternating current is found using the equations for the alternating currents and voltages that are produced. The formula for power dissipated over a cycle is found out in terms of the work done per unit time and then integrated over a time period to find the average power that is dissipated over a cycle.

Complete step by step answer:
The above problem revolves around the concept of the power dissipated in an ac circuit. In order to find the average power we first need to know the concept behind power and alternating currents. Power is defined as the rate at which the work is done by a source of emf or voltage supply in maintaining an electric current throughout the circuit. The power is a measure of the heat that is dissipated due to the production of current when there is a voltage supply connected across its circuit elements.

Alternating current, unlike direct current, changes or varies continuously with time and this is represented through cycles where the variation of current over a period of time is measured since it changes periodically at every instant. Since this is periodic in nature, the alternating currents are said to be represented by sinusoidal waves wherein its direction reverses periodically as well. Periodic waves, as we know, are indicated using sine and cosine functions. We are asked to prove that the average power dissipated over a cycle to be $\dfrac{1}{2}{i^2}R$. This is possible only when we consider a pure resistor circuit which means that the circuit consists only of a resistor as its circuit element.

Since both voltage and current are said to be alternating in nature, there must exist a phase difference between them and this phase difference varies from element to element, that is, it is dependent on the circuit element used. In an ac circuit, the current may lag behind or lead the voltage depending on the circuit through which it flows through. This is found out using phasor diagrams, that is, a rotating vector used to represent current and voltage.

In a pure resistor circuit there is said to be no phase difference between the current and voltage values and hence the voltage and the current produced are said to be in the same phase. Since here we are considering a purely resistor circuit we consider equations for current and voltage wherein there is no phase difference between them.
Hence the equation for alternating current is given in the question to be:
$i = {i_m}\sin \omega t$ -------($1$)
This shows the variation of current with time wherein the equation represents the instantaneous value of current, that is, the value of current at a particular instant of time $t$. Here, ${i_m}$ denotes the peak current or the maximum value of current.

Since there is an alternating current which is produced there will be a voltage supply producing this current. Hence we now determine a similar equation for the instantaneous value of voltage at a particular instant of time $t$. This equation for voltage is given to be:
$v = {v_m}\sin \omega t$ --------($2$)
Since there is some work that is being done in order to dissipate some heat due to the flow of charges through the circuit element, that is, resistor we first determine an equation for the work done in order to determine the average power.

The work done is first considered over a small interval of time and then integrated over an entire time interval to find the total or the average power that is dissipated over a cycle.
Hence the work done over a small interval $dt$ is:
$dW = Pdt$ -------($3$)
Where, $P$ represents the instantaneous power.
We now determine the instantaneous value of power from the equations ($1$) and ($2$). We know that the general formula for power is
$P = VI$
By using the same concept we find the instantaneous value of power to be:
From equations ($1$) and ($2$) we get:
$P = v \times i$
$\Rightarrow P = {v_m}\sin \omega t \times {i_m}\sin \omega t$
By solving further we get:
$\Rightarrow P = {v_m}{i_m}{\sin ^2}\omega t$
We substitute the above value in the equation ($3$) to get:
$dW = \left[ {{v_m}{i_m}{{\sin }^2}\omega t} \right]dt$

We now use trigonometry, to apply the double angle formula to get the value of sine in terms of cos. The double angle formula is given as:
$\cos 2\theta = 1 - 2\sin \theta$
$\Rightarrow 2\sin \theta = 1 - \cos 2\theta$
Here, the $\theta$ value is taken as $\omega t$. Putting $\theta = \omega t$ we get:
$\Rightarrow 2\sin \omega t = 1 - \cos 2\omega t$
$\Rightarrow \sin \omega t = \dfrac{{1 - \cos 2\omega t}}{2}$
Hence we substitute the above value in terms of cos in the work done equation to get:
$dW = {v_m}{i_m}\left[ {\dfrac{{1 - \cos 2\omega t}}{2}} \right]dt$ ---------($4$)
We know that the average power is given by the formula:
$P = \dfrac{W}{T}$
Where, $T$ is the time period that is considered to be a period of one cycle.
$P = \dfrac{{dW}}{T}$

In order to find the work done over the entire time period we must integrate it over this time period. Hence we get:
$P = \dfrac{1}{T}\int\limits_0^T {dW}$
We now substitute equation ($4$) in the above power equation:
$P = \dfrac{1}{T}\int\limits_0^T {{v_m}{i_m}\left[ {\dfrac{{1 - \cos 2\omega t}}{2}} \right]dt}$
Constants are taken out of the integral. Hence we get:
$P = \dfrac{{{v_m}{i_m}}}{{2T}}\int\limits_0^T {\left[ {1 - \cos 2\omega t} \right]dt}$
We now perform the integration with respect to time and expanding the brackets.
$P = \dfrac{{{v_m}{i_m}}}{{2T}}\left[ {\int\limits_0^T {dt - \int\limits_0^T {\cos 2\omega tdt} } } \right]$
By solving further we get:
$\Rightarrow P = \dfrac{{{v_m}{i_m}}}{{2T}}\left[ {\left( {T - 0} \right) - \int\limits_0^T {\cos 2\omega t} } \right]$
$\Rightarrow P = \dfrac{{{v_m}{i_m}}}{{2T}}\left[ {T - \left[ {\dfrac{{\sin 2\omega t}}{{2\omega t}}} \right]_0^T} \right]$ [Since,$\int {\cos 2\theta = \dfrac{{\sin 2\theta }}{2}}$]

We know that, $\sin 0 = 0$ and also the sine function over a time period is also said to be zero since sine is in the form $\sin 2\theta$ which will be equivalent to zero whatever be the value of $\theta$. Hence we get:
$\Rightarrow P = \dfrac{{{v_m}{i_m}}}{{2T}}\left[ {T - 0} \right]$
$\Rightarrow P = \dfrac{{{v_m}{i_m}T}}{{2T}}$
Cancelling out the common terms we get:
$\Rightarrow P = \dfrac{{{v_m}{i_m}}}{2}$ -------($5$)
We now apply the ohm’s law formula to get the above equation in terms of resistance of the resistor considered in the circuit. Let $R$ denote the value of resistance. Hence we know that: $V = IR$
Similarly,
${v_m} = {i_m}R$
Substituting this value in the equation ($5$) we get:
$\Rightarrow P = \dfrac{{{i_m}R \times {i_m}}}{2}$
$\therefore P = \dfrac{{{i^2}_mR}}{2}$
Hence proved.

Hence the total power or the average power dissipated over a cycle is said to be $\dfrac{1}{2}{i^2}_mR$.

Note: There may be a common error that is made for the above problem where instantaneous power is calculated instead of the average power which is wrong. The instantaneous power that is calculated must be integrated over a time period. The phase difference between the voltage and current values must also be taken into consideration.