Question

an alternating current is given by the expression i=50sin628t.find maximum value of current after t=0.00625sec ,frequency of current,rms value of current,time taken by the current to rdach a value of 20Amp.​

Answer 1

Answer:

we know that

i=Im sinwt

given i=50sin628t

by comparing

Answer 2

The answers are as follows-

Maximum value of current at 0.00625 is -35.28A.

Maximum value of current at 0.00625 is -35.28A. Rms value of the current= 35.35 A

Maximum value of current at 0.00625 is -35.28A. Rms value of the current= 35.35 ATime taken is 0.411 second

GIVEN

Expression of AC Current, I = 50sin628t

Time = 0.00625 second

TO FIND

• The maximum value of the current.
• RMS value of the current.
• Time taken by the current to reach a value of 20 Amp.

SOLUTION

We can simply solve the above problem as follows-

We are given,

I = 50sin628t. (equation 1)

The standard equation for alternating current is represented as -

I = I₀sinωt (equation 2)

Where,

I₀ = maximum value of the current

Comparing equations 1 and 2, we get

I₀ = 50 Amp

At, t = 0.00625 sec

Putting the value of t in equation 1

I = 50 sin628 × 0.00625

I = - 35.28 A

Hence, The maximum value of the current is 50 Amp.

Amd, maximum value of current after 0.00625 second is -35.28 A

RMS value of the current.

Let the RMS value of the current be, Iᵣₘₛ

We know that,

Iᵣₘₛ = I₀/√2

Where,

I₀= maximum value of the current= 50 A

Therefore,

Iᵣₘₛ = 50/√2

Iᵣₘₛ = 35.35 A

Time taken by the current to reach a value of 20 Amp.

We know that,

I = I₀sinωt

Putting the values in the above formula we get,

20 = 50sin628t

628t = sin⁻¹ (2/5 )

t = 0.4115 second.

Hence, The time taken is 0.411 second .

The answers are as follows-

1. The answers are as follows- Maximum value of current at 0.00625 is -35.28A.

1. Rms value of the current= 35.35 A

1. Time taken is 0.411 second

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