Question

(3) The values of the instantaneous currents
in the branches of a parallel circuit are as
follows: i1 = 5 sin 346 t; i2 = 10 sin (346t +
π/4); i3 = 7.5 sin (346 t + π/2); i4 = 8 sin (346
t - π/3)Express the resultant line current in
the same form as the original expression
and determine the r.m.s.value and the
frequency of this current.​

Answer 1

Answer:

inet = 18 sin (346 t + 64⁰) , rms value of i = 12.73 A , frequency = 0.96 Hz

Explanation:

i1 = 5sin346t , i2 = 10sin(346t+45⁰) , i3 = 7.5sin(346t+90⁰) , i4 = 8sin(346t-60⁰)

let 346t be x

i1 = 5sinx

i2 = 10sin(x+45⁰) = 10(sinxcos45⁰+sin45⁰cosx) = 5(1.414)sinx + 5(1.414)cosx

i3 = 7.5sin(x+90⁰) = 7.5cosx

i4 = 8sin(x-60⁰) = 8(sinxcos60⁰-sin60⁰cosx) = 4sinx - 4(1.73)cosx

inet = i1+i2+i3+i4

= (5+5(1.414)+4)sinx + (5(1.414)+7.5-4(1.73))cosx

= 16.07sinx + 7.64cosx

=√((16.07)²+(7.64)²)sinx(16.07/√((16.07)²+(7.64)²))

+ √((16.07)²+(7.64)²)cosx(7.64/√((16.07)²+(7.64)²))

let tanα = 16.07/7.64

therefore sinα = 16.07/√(16.07)²+(7.64)²)

cosα = 7.64/√((16.07)²+(7.64)²)

on simplifying

inet = √((16.07)²+(7.64)²)(sinxsinα+cosxcosα)

=17.79cos(x-α) = 17.79sin(x+α)

=18sin(x+arctan(16.07/7.64))

=18sin(x+64.5⁰)

=18sin(346t+64⁰)

comparing with eq of inet

Io=18 , Irms = 18/1.414 = 12.73 A

w = 346Hz , f = 346/2(180⁰) = 0.96 Hz