Alternating Voltage Applied to a Resistor

The applied alternating voltage is given by,  

$V = V_0 \sin \omega t$

Let ( I ) be the current in the circuit at any instant ( t ).

As per Ohm's Law, potential difference across the resistor,  

$V = IR$

or  

$I = \frac{V}{R}$

Using eqn. (i), we get,  

$I = \frac{V_0 \sin \omega t}{R}$

or  

$I = I_0 \sin \omega t$

where,  

$I_0 = \frac{V_0}{R}$

is the peak value of alternating current.

If a graph for voltage across the resistor and current in the resistor is plotted with respect to time ( t ), it is observed that voltage and current have zero, maximum and minimum values at the same times respectively. Moreover, comparison of eqn. (i) and (ii) shows that the current and voltage across the resistor are in phase with each other.

Phasor diagram for pure resistive circuit shows that the phase difference between current ( I ) and voltage ( V ) is zero.

Instantaneous Power : 

The instantaneous power dissipated in resistor as per Joule’s heating effect is given by  

$P = I^2 R = (I_0^2 \sin^2 \omega t) R$

Average Power : 

$P_{av} = \text{average of } I^2 R$

$P_{av}= \text{average of } I_0^2 R \sin^2 \omega t$

Since,  

$\sin^2 \omega t = \tfrac{1}{2}(1 - \cos 2\omega t)$

Average of ( $\cos 2\omega t$) over a full cycle = 0.  

Therefore,  

$\langle \sin^2 \omega t \rangle = \tfrac{1}{2}$

Substituting,  

$P_{av} = \tfrac{1}{2} I_0^2 R$

Now, 

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

So, 

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

RMS (Root Mean Square) Values : 

To express AC power in the same form as DC power ( $P = I^2 R$), a special value of current is defined and used. It is called root mean square (rms) or effective current and is denoted by ( $I_{rms}$ ) or ( I).

It is defined by  

$I_{rms} = \sqrt{\bar{i^2}} = \sqrt{\tfrac{1}{2} i_0^2} = \frac{i_0}{\sqrt{2}} = 0.707 \, i_0 \tag{7.6}$

In terms of ( I ), the average power is  

$P = \bar{P} = \tfrac{1}{2} i_0^2 R = I^2 R \tag{7}$

Similarly, rms voltage or effective voltage is defined by  

$V_{rms} = \frac{v_0}{\sqrt{2}} = 0.707 \, v_0 \tag{7.8}$

It is customary to measure and specify rms values for AC quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of  

$v_0 = \sqrt{2} V = (1.414)(220 \, V) = 311 \, V$

In fact, the rms current ( I ) is the equivalent DC current that would produce the same average power loss as the alternating current.

$P = \frac{V^2}{R} = VI \quad \text{(since } V = IR \text{)}$