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c) Three AC voltages are as follows:
e1 = 80 sin ωt volts;
e2 = 60 sin (ωt + π/2) volts;
e3 = 100 sin (ωt – π/3) volts.
Find the resultant e of these three voltages and express it in the form
Em sin (ωt ± φ). [5 MARKS]
When this resultant voltage is applied to a circuit consisting of a 10-Ω resistor and a capacitance of 17.3 Ω reactance connected in series, find an expression for the instantaneous value of the current flowing, expressed in the same form. [4 MARKS]

Respuesta :

Answer:

E = 132.69 sin(ωt -11.56)

i(t) = 6.64 sin (ωt +48.44) A

Explanation:

given data

e1 = 80 sin ωt volts                       80 < 0

e2 = 60 sin (ωt + π/2) volts          60 < 90

e3 = 100 sin (ωt – π/3) volts        100 < -60

solution

resultant will be  = e2 + e2 + e3

E =  80 < 0 + 60 < 90 + 100 < -60

[tex]\bar E[/tex] = 80 + j60 + 50 - j50[tex]\sqrt{3}[/tex]

[tex]\bar E[/tex] = 130 + (-j26.60)

[tex]\bar E[/tex] = 132.69  that is less than -11.56

so

E = 132.69 sin(ωt -11.56)

and

as we have given the impedance

z = (10-j17.3)Ω

z = 19.982 < -60

and

i(t) = [tex]\frac{132.69}{19.982}[/tex] sin(ωt -11.56 + 60)  

i(t) = 6.64 sin (ωt +48.44) A

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