Question

Problem 1 (10 Points) A voltage source is given by v = 2cos(60t + 45o ). What is the amplitude of the voltage source? What is the frequency of the voltage source in radians/second? What is the phase of the voltage source? What is the period of the voltage source? What is the frequency of the voltage source in Hz? What is the RMS voltage of the voltage source?

Answer 1

Answer:

The amplitude of the voltage source is 2.

The frequency of the voltage in rad/s is 60.

The phase of this voltage source is 45 degrees.

The period is $T = \frac{\pi}{30}s$

The frequency in Hz is $f = \frac{30}{\pi}Hz$

The RMS voltage of the voltage source is 1.41V.

Explanation:

Any sinodal function has amplitude 1 when not multiplied by a coefficient c. Here we have $v = c/cos{60t + 45}$, in which $c = 2$. So, the amplitude is 2.

For a function $y = /cos{mx + a}$, the period is given by the following formula: $T = \frac{2\pi}{m}$. For this problem, we have a period of:

$T = \frac{2\pi}{60} = \frac{\pi}{30}s$

In voltage source in the format $V = c\cos{mt + a}$, the phase is a. So, for this voltage source, the phase is 45 degrees.

The frequency, in Hz, is given by the following function.

$f = \frac{1}{T}$

So

$f = \frac{1}{\frac{\pi}{30}} = \frac{30}{\pi}Hz$

To convert to rad/s, we solve the following rule of three.

1Hz - $2\pi rad/s$

$\frac{30}{\pi}Hz - x rad/s$

$x = 60 rad/s$

The RMS voltage is given by the following formula:

$V_{RMS} = \frac{V_{peak}}{\sqrt{2}}$

This voltage source has amplitude 2, so $V_{peak} = 2$

$V_{RMS} = \frac{V_{peak}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = 1.41$