Question

The current and voltage in circuit is given by i=3.5sin(628t+30°) v=28sin(628t+30°)vol
1. Time period of current
2. Phase difference between voltage and current

Answer 1

Answer:

An alternating voltage e=200 sin 314t is applied to a device which offers an ohmic resistance of 20 ohm to the flow of current in one direction, while preventing the flow of current in opposite direction. Calculate rms value, average value and form factor for the current over one cycle.

Solution:

The instantaneous voltage applied to the rectifying device is given by the expression

e = 200 sin 314 t

Maximum value of applied voltage,

Emax = Coefficient of the sine of time angle = 200 volts

Resistance of rectifying device, R = 20 ohm

Maximum value of half-wave rectified alternating current,

I_{max} = \dfrac{E_{max}}{R} = \dfrac{200}{20} = 10\ amperes

RMS value of half-wave rectified alternating current,

I_{rms} = \dfrac{I_{max}}{2} = \dfrac{10}{2} = 5\ amperes Ans.

Average value of the half wave rectified alternating current,

I_{av} = \dfrac{I_{max}}{\pi} = \dfrac{10}{\pi} = 3.18\ amperes Ans.

Form factor of the half-wave rectified alternating current

= \dfrac{I_{rms}}{I_{av}} = \dfrac{5}{3.18} = 1.57 Ans.

_________________

hope it is helpful

Answer 2

Time period of current is 0.01 s and phase difference is 0° .

Given :

Equation of current ,  i=3.5 sin(628t+30°) .

Equation of voltage , v=28sin(628t+30°)  .

Ideal equation , $A\ sin(\omega t+\phi)$ .

We need to find the time period of current .

Comparing equation of current with ideal equation :

$\omega=628\ rad/s$ .

We know ,

Time period is given by :

$T=\dfrac{2\pi}{\omega}$

Putting value in above equation :

We get ,

$T=\dfrac{2\pi}{628}\\\\T=0.01 \ s$

Now , phase difference between voltage and current source is zero because they both are in same phase .

Hence , this is the required solution .

Learn More :

Alternating current

https://brainly.in/question/11615655