Question

The emf phasor in FIGURE EX32.2 is shown at $t=15ms$.

a. What is the angular frequency $\omega$? Assume this is the first rotation.

b. What is the instantaneous value of the emf?

Short answer

The angular frequency $\omega$is $244rad/s$and instantaneous value of the emf is$-10V$.

Step-by-step solution

Part (a) Step 1: Given information

We are given that time $t=15ms$, $E_o=12V$and $\varnothing ={150}^{\circ }$(as the phasor rotates anticlockwise towards X-axis)

Part (a) Step 2: Calculation

As we know that-

$E=E_0\cos (\omega t+\varnothing )$

$$\begin{gathered} \omega t={180}^{\circ }+\varnothing \\ \omega t={180}^{\circ }+{30}^{\circ } \\ \omega t={210}^{\circ } \end{gathered}$$ (as phasor rotates anticlockwise)

Now converting degrees into radian

${210}^{\circ }={210}^{\circ }\frac{\mathrm{\pi rad}}{{180}^{\circ }}=\frac{7\mathrm{\pi rad}}{6}$

Now,

$$\begin{gathered} \omega t=\frac{7\mathrm{\pi rad}}{6} \\ \omega (15\times {10}^{-3})=\frac{7\mathrm{\pi rad}}{6} \\ \omega =244rad/s \end{gathered}$$

Part (b) Step 1: Given information

We are given that-

$$\begin{gathered} E_0=12V \\ \varnothing ={150}^{\circ } \end{gathered}$$

Part (b) Step 2: Calculation

As we know that peak value$E_0=12V$ will occur when $\omega t=0$

So now in the expression,

$$\begin{gathered} E=E_0\cos (\omega t+\varnothing ) \\ E=12V\cos (0+{150}^{\circ }) \\ E=-10V \end{gathered}$$