Alternating Voltage Applied to a Resistor - Class 12 Physics

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,  

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

$I_0$ is the peak value of alternating current.

Alternating Voltage Applied to a Resistor

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

Alternating Voltage Applied to a Resistor

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.

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