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AC Supply to Pure Resistor

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)

purely resistive circuit

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.

purely resistive circuit

Power calculation

The instantaneous power p(t) is defined by the product of instantaneous voltage v(t) and instantaneous current i(t).

purely resistive circuit

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}

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