Convert the following pairs of voltage and current waveforms to phasor form. Each pair of waveforms corresponds to an unknown element. Determine whether each element is a resistor, a capacitor, or an induotor, and compute the value of the corresponding parameter R, C, or L. 2.

a. v(t) = 20 cos(400t +30), i (t) = 4 sin(400t – 120)
b. b. (t) = 9 cos(900t - 60), i (t) = 4 sin(900t + 280)
c. c. vt) - 13 cos(250+ + 60), 1 (t) = 7 sin(250t + 240)

a) V = 20 ∠30⁰    ,    I = 4 ∠-210⁰    Z inductive    L = 0,0125 H

b) V = 9∠-60⁰      ,    I = 4 ∠ 190⁰    Z capacitive C = 4,94 *10⁻⁴ F

c) V = 13 ∠240⁰   ,    I = 7 ∠ 150⁰    Z Inductive  L = 0,0074 H

Explanation:

a) v(t) = 20 cos (400*t + 30 )

Phasor form    V = 20 ∠30⁰

i(t) = 4 sin (400*t - 120)

First we need to transform 4sin( 400t - 120 ) as  function cosine

we know that  sin ( x + 90 )  =  cos x

Then  sin ( 400*t -120 )  = cos ( 400*t  - 120 -90 )  = cos ( 400t - 120 - 90)

Phasor form  I = 4 ∠-210⁰

To have the impedance nature we compute

Z = V / I      ⇒  Z = 20 ∠30⁰ / 4  ∠-210⁰    Z = 5 ∠-180⁰

We notice that  voltage advances the current then we are in presence of an inductive impedance

5 = wl      ⇒  5  = 400 *L       ⇒  L  =    0,0125 H        

b) v(t) = 9 cos ( 900t - 60 )

V = 9∠-60⁰

i(t)  = 4 sin ( 900t + 280 )    ⇒  i(t) = 4 cos ( 900t + 280 - 90)

i(t) = 4 cos (900t + 190 )    ⇒  I = 4 ∠ 190⁰

Z = V/I    ⇒  Z = 9∠-60⁰ / 4  ∠ 190⁰    Z = 2,25 ∠-250

In this case the current advances the voltage. Impedance capacitive

1/wc  = 1/ 900*C       1/wc = Z   ⇒ 2,25 = 1/ 900*C

2,25*900 = 1/C     ⇒  2025 =1/C     ⇒  C = 4,94 *10⁻⁴ F

c) v(t) = - 13 cos ( 250t + 60 )

v(t) = 13 cos ( 250t + 60 +180 )    ⇒ v(t) = 13 cos ( 250t +240)

Phasor Form

V = 13 ∠240⁰

i(t) = 7 sin (250t + 240 - 90)  ⇒  i(t) = 7 sin (250t + 150)

Phasor Form  I = 7  ∠150⁰

Z = 13∠240⁰ / 7 ∠150⁰    ⇒  Z = 1,86 ∠ 90⁰

Voltage advances the current then the impedance is inductive

wl = 250L     250 L = 1,86     L  = 1,86/250     L = 0,0074 H

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