Calculating the Phase Angle Between AC Voltages

To find the phase angle between two sinusoidal voltages, both voltages must be expressed in the same standard form, such as $A \cos(\omega t + \phi)$ or $A \sin(\omega t + \phi)$. Let's convert both given voltages, $v_1$ and $v_2$, into the standard cosine form $A \cos(\omega t + \phi)$.

The given voltages are:

• $v_1 = -10 \cos (\omega t + 50^\circ)$
• $v_2 = 12 \sin (\omega t – 10^\circ)$

Let's convert $v_1$ first. We have a negative cosine term. We can use the trigonometric identity $-\cos(\theta) = \cos(\theta + 180^\circ)$.

Let $\theta = \omega t + 50^\circ$. Then,

$$v_1 = -10 \cos (\omega t + 50^\circ) = 10 \cos ((\omega t + 50^\circ) + 180^\circ)$$

$$v_1 = 10 \cos (\omega t + 230^\circ)$$

So, the phase angle for $v_1$ in the standard cosine form is $230^\circ$. We can also express this as $230^\circ - 360^\circ = -130^\circ$. Let's use $230^\circ$ for now.

Next, let's convert $v_2$ to the standard cosine form. We have a sine term. We can use the trigonometric identity $\sin(\theta) = \cos(\theta - 90^\circ)$.

Let $\theta = \omega t – 10^\circ$. Then,

$$v_2 = 12 \sin (\omega t – 10^\circ) = 12 \cos ((\omega t – 10^\circ) - 90^\circ)$$

$$v_2 = 12 \cos (\omega t – 100^\circ)$$

So, the phase angle for $v_2$ in the standard cosine form is $-100^\circ$.

Now that both voltages are in the same standard form ($A \cos(\omega t + \phi)$), we can find the phase angle between them by taking the difference of their phase angles.

Phase difference = Phase angle of $v_1$ - Phase angle of $v_2$

Phase difference = $230^\circ - (-100^\circ)$

Phase difference = $230^\circ + 100^\circ$

Phase difference = $330^\circ$

The phase difference is $330^\circ$. However, phase angles are typically expressed between $-180^\circ$ and $+180^\circ$ or between $0^\circ$ and $360^\circ$. A difference of $330^\circ$ is equivalent to $330^\circ - 360^\circ = -30^\circ$. The magnitude of the phase difference is $30^\circ$.

Alternatively, if we calculate Phase angle of $v_2$ - Phase angle of $v_1$:

Phase difference = $-100^\circ - 230^\circ = -330^\circ$. This is equivalent to $-330^\circ + 360^\circ = 30^\circ$.

The magnitude of the phase angle between $v_1$ and $v_2$ is $30^\circ$. A positive difference means the first signal leads the second, and a negative difference means it lags. The question asks for "the phase angle", which usually refers to the magnitude or the relative difference. In this case, the magnitude is $30^\circ$.

Let's also check the conversion to the sine form $A \sin(\omega t + \phi)$.

Convert $v_1 = -10 \cos (\omega t + 50^\circ)$ to sine form using $-\cos(\theta) = \sin(\theta - 90^\circ)$.

$$v_1 = -10 \cos (\omega t + 50^\circ) = 10 \sin ((\omega t + 50^\circ) - 90^\circ)$$

$$v_1 = 10 \sin (\omega t - 40^\circ)$$

Phase angle for $v_1$ in sine form is $-40^\circ$.

Convert $v_2 = 12 \sin (\omega t – 10^\circ)$. This is already in sine form.

Phase angle for $v_2$ in sine form is $-10^\circ$.

Phase difference = Phase angle of $v_2$ - Phase angle of $v_1$

Phase difference = $-10^\circ - (-40^\circ)$

Phase difference = $-10^\circ + 40^\circ$

Phase difference = $30^\circ$

Using either method (converting to cosine or sine form), the phase angle difference between the two voltages is $30^\circ$.

The correct option is the one stating $30$ degrees.