Answer:
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
