Answer to Question #294298 in Electric Circuits for Sagar

Question

The instaneous value of voltage in an a. c circuit at any time t seconds is given by v=100sin(50πt-0.523) V. Find

1:The peak to peak voltage, the frequency and the phase angle

2:the voltage when t=0

3:the voltage when t=8ms

4:the times in the first cycle when the voltage is 60V and -40V

5:the first time when the voltage is maximum

Answer 1

Part 1

peak to peak voltage $=100×2=200V$

$2\pi\>f=50\pi\implies\>f=25Hz$

Phase angle$=0.523$ radians ; lagging

Part 2

$V=100\>Sin(-0.523)$

$=100×-0.49948$

$=-49.948V$

Part 3

$V=100$ Sin $(50\pi×8×10^{-3}-0.523)$

$=100×0.6697$

$=66.97V$

Part 4

$60=100$ Sin $(50\pi\>t-0.523)$

$50\pi\>t-0.523=$ Sin$^{-1}\frac{60}{100}$

$50\pi\>t-0.523=0.6435$

$t=7.426ms$

$-40=100$ Sin $(50\pi\>t-0.523)$

$50\pi\>t-0.523=$ Sin$^{-1}(\frac{-40}{100})$

$50\pi\>t-0.523=-0.4115$

$t=0.7097ms$

Part 5

$100=100$ Sin$(50\pi\>t-0.523)$

$50\pi\>t-0.523=$ Sin$^{-1}1$

$50\pi\>t-0.523=1.5708$

$t=13.33ms$