Question

Show that the power dissipated in a resistor connected to an \(\mathrm{AC}\) power source with a frequency \(\omega\) oscillates with a frequency \(2 \omega\).

Short answer

Answer: The frequency of oscillation in the power dissipation of a resistor connected to an AC source is 2ω, where ω is the angular frequency of the AC voltage source.

Step-by-step solution

Write down the expression for the AC voltage source

An AC voltage source oscillates sinusoidally according to the equation: \(V(t) = V_0 \cos(\omega t)\) where \(V_0\) is the amplitude of the voltage, \(t\) is time, and \(\omega\) is the angular frequency of oscillation.

Find the expression for the current flowing through the resistor

Using Ohm's Law, the current can be calculated by dividing the voltage across the resistor by its resistance: \(I(t) = \frac{V(t)}{R}\) Substituting the equation for AC voltage source: \(I(t) = \frac{V_0 \cos(\omega t)}{R}\)

Write the expression for power dissipation in the resistor

The power dissipated in a resistor can be calculated using the equation: \(P(t) = I(t) \times V(t)\) Substituting the equations found in Steps 1 and 2: \(P(t) = \left(\frac{V_0 \cos(\omega t)}{R}\right) \times (V_0 \cos(\omega t))\) \(P(t) = \frac{V_0^2 \cos^2(\omega t)}{R}\)

Use Trigonometric identity

To analyze the frequency of oscillation in the power expression, we'll use the trigonometric identity for the double angle of cosine: \(\cos^2(x) = \frac{1+\cos(2x)}{2}\) Applying this to our equation for power: \(P(t) = \frac{V_0^2}{2R} \left( 1 + \cos(2\omega t)\right)\)

Identify the frequency of oscillation in the power expression

From the above equation, we can see that the power dissipation oscillates with respect to the term \(\cos(2\omega t)\). This term has a frequency of \(2\omega\). Therefore, the power dissipated in a resistor connected to an AC source oscillates with a frequency of \(2\omega\).