Question

Q. An alternating voltage is given by,
e=6 sin 314 t. find
i. the peak value
ii. frequency
iii. time period and
iv. instantaneous value at time t= 2 ms​

Answer 1

Answer:

1. 6. is the Peak value since Sine is Max =1

2. w = 314

so f = 314/2 Pi =50

3. T = 1/f =;1/50=0.02 sec

4. e at 2 = 6x sine 628 nearly 6 sin 270 = nearly -6

Answer 2

Given:

• E = 6 sin (314 t)

To find:

• Peak value
• Frequency
• Time period
• Instànt value at t = 2 sec

Calculation:

$e = 6 \sin(314t)$

• We know π = 3.14, we can rearrange as :

$e = 6 \sin(100\pi t)$

• Max value of sine function is +1 , so Max value of voltage is :

$e_{peak} = 6 \times 1 = 6 \: volt$

Now, Frequency is :

$\implies f = \dfrac{ \omega}{2\pi}$

$\implies f = \dfrac{ 100\pi}{2\pi}$

$\implies f = 50 \: hz$

So, time period will be :

$\implies t = \dfrac{1}{f}$

$\implies t = \dfrac{1}{50}$

$\implies t = 0.02 \: sec$

So, instànt value at t = 2 sec is :

$e = 6 \sin(100\pi t)$

$\implies e = 6 \sin(100\pi \times 2)$

$\implies e = 6 \sin(200\pi )$

$\implies e =0 \: volt$

Hope It Helps