Question

The emf phasor in FIGURE EX32 .1 is shown at$t=2.0ms$.

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

b. What is the peak value of the emf?

Short answer

The angular frequency $\omega =1200rad/s$ and peak value of emf is$E_0=71V$.

Step-by-step solution

Part (a) Step 1: Given information

We are given that

$$\begin{gathered} t=2ms \\ \varnothing ={180}^{\circ }-{150}^{\circ }={45}^{\circ } \end{gathered}$$

(as the phasor rotates anticlockwise towards X-axis)

Part (a) Step 2: Calculation

As we know that-

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

Now converting 135 degree into radian

${135}^{\circ }={135}^{\circ }\times \frac{\mathrm{\pi rad}}{{180}^{\circ }}=\frac{3\mathrm{\pi }}{4}$

From here,

$$\begin{gathered} \omega t=\frac{3\mathrm{\pi }}{4} \\ \omega (2\times {10}^{-3})=\frac{3\mathrm{\pi }}{4} \\ \omega \approx 1200rad/s \end{gathered}$$

Part (b) Step 1: Given information

We are given that

$E=50V$when $\omega t=0$and $\varnothing ={45}^{\circ }$.

Part (b) Step 2: Calculation

Now from the expression-

$$\begin{gathered} E=E_0\cos (\omega t+\varnothing ) \\ 50V=E_0\cos (0+{45}^{\circ }) \\ {E}_0=\frac{50V}{\cos \left({45}^{\circ }\right)} \\ {E}_0=71V \end{gathered}$$