Response of Resistor to a Sinusoidal Voltage

Now that we are familiar with the characteristics of the derivative of a sinusoidal function, we can investigate the response of the basic elements $R$, $L$, and $C$ to a sinusoidal voltage or current.
For power-line frequencies and frequencies up to a few hundred kilo-hertz, resistance is, for all practical purposes, unaffected by the frequency of the applied sinusoidal voltage or current. For this frequency region, the resistor R of [Fig. 1] can be treated as a constant, and Ohm's law can be applied as follows. For $v = V_m \sin wt$,
$$i = {V \over R}={V_m \sin wt \over R} = I_m \sin wt$$
$$\bbox[10px,border:1px solid grey]{I_m = {V_m \over R}} \tag{1}$$
Fig. 1: Determining the sinusoidal response for a resistive element.
Fig. 2: The voltage and current of a resistive element are in phase.
In addition, for a given i,
$$v=iR = (I_m \sin wt)R = V_m \sin wt$$
$$ \bbox[10px,border:1px solid grey]{V_m = ImR} \tag{2}$$
A plot of $v$ and $i$ in [Fig. 2] reveals that
for a purely resistive element, the voltage across and the current through the element are in phase, with their peak values related by Ohm's law.
Example 1: The voltage across a resistor is indicated. Find the sinusoidal expression for the current if the resistor is $10 Ω$. Sketch the curves for v and i.
a. $v = 100 \sin 377t $
b. $v = 25\sin (377t + 60)$
Solution: a. Eq. (1): $I_m = V_m / R = {100V \over 10Ω} = 10 A$
(v and i are in phase), resulting in
$$i = 10 \sin 377t$$
The curves are sketched in [Fig. 3].
Fig. 3: Example 1(a).
b. Eq. (1): $I_m = V_m / R = {25V \over 10Ω} = 2.5 A$
(v and i are in phase), resulting in
$$i = 2.5 \sin (377t + 60)$$
The curves are sketched in [Fig. 4].
Fig. 4: Example 1(b).

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