After reading this AC supply to pure Resistor topic of electric or network circuits, you will understand the theory, waveforms, phasor, formula, & also voltage, current, power calculation.
Voltage and Current Relationship
Let us consider a circuit having resitance R supplied by a ac source voltage v(t) as shown in Figure 1 is
v(t) = {V_m}\sin \omega t ….(1)
The current passes throgh resistance R will be
i(t) = \frac{{v(t)}}{R} = \frac{{{V_m}}}{R}\sin \omega t = {I_m}\sin \omega t
Thus,
i(t) = {I_m}\sin \omega t ….(2)
Phasor Diagram
The phasor representation of Equations (1) and (2), is shown in Figure 2, can be written and relate in phasor terms as
{\mathbf{I}} = \frac{{\mathbf{V}}}{R}
or,
{\mathbf{V}} = {\mathbf{I}}R
Equations (1) and (2) shows that both current and voltage are in phase so waveforms will look as shown in Figure 3.
Power calculation
The instantaneous power p(t) is defined by the product of instantaneous voltage v(t) and instantaneous current i(t).
The instantaneous power waveform as shown in Figure 3. And according to the definition, instantaneous power p(t) write as
p(t) = v(t)i(t) ….(3)
Using Equation (3), the average power is defined as
{P_{av}} = \frac{1}{T}\int\limits_0^T {p(t)dt}
Also,
{P_{av}} = \frac{1}{T}\int\limits_0^T {v(t)i(t)dt}
= \frac{1}{T}\int\limits_0^T {\left( {{V_m}\sin \omega t} \right)\left( {{I_m}\sin \omega t} \right)dt}
= \frac{{{V_m}{I_m}}}{T}\int\limits_0^T {{{\sin }^2}\omega tdt}
= \frac{{{V_m}{I_m}}}{{2T}}\int\limits_0^T {\left( {1 – \cos 2\omega t} \right)dt}
= \frac{{{V_m}{I_m}}}{2}
= \frac{{{V_m}}}{{\sqrt 2 }}\frac{{{I_m}}}{{\sqrt 2 }}
= {V_{rms}}{I_{rms}}
Thus,
{P_{av}} = {V_{rms}}{I_{rms}}.
Also, Figure 3 shows that the frequency of instantaneous power {f_p} is twice the frequency of instantaneous voltage {f_v} or the frequency of instantaneous current {f_i} as
{f_p} = 2{f_i}{\text{ }}or{\text{ }}2{f_v}