AC Through Pure Ohmic Resistance Alone
The circuit is shown in Figure (A) is explaining AC Through Pure Ohmic Resistance Alone. Let the applied voltage be given by the equation.
v = Vm sin θ = Vm sin ωt ………………………….(i)
Let R = ohmic resistance ;
i = instantaneous current
Obviously, the applied voltage has to supply ohmic voltage drop only. Hence for equilibrium
v = iR;
Putting the value of ‘v’ from above,
we get,
Vm sin ωt = iR;
Current ‘i’ is maximum when sin ωt is unity
∴ Im = Vm/R Hence, equation (ii) becomes,
i = Im sin ωt …—————————-(iii)
Comparing (i) and (ii), we find that the alternating voltage and current are in phase with each other as shown in Figure (B).
Instantaneous power, p = vi = VmIm sin2ωt …………………………….. (Figure (C))
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Power consists of a constant part 
and a fluctuating part cos2ωt of frequency double that of voltage and current waves. For a complete cycle. the average value of cos 2ωt is zero.
Hence, power for the whole cycle is
or P = V × I watt
where V = r.m.s. value of applied voltage.
I = r.m.s. value of the current.
It is seen from Figure (C) that no part of the power cycle becomes negative at any time. In other words, in a purely resistive circuit, power is never zero. This is so because the instantaneous values of voltage and current are always either both positive or negative and hence the product is always positive.
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