jawapan ca2 lab

CALCULATION

By using capacitor value = 0.01µF and measured resistance = 0.987kΩ :

Z_c = -j

wC

= -j1591.6Ω

Z_T = R//Z_c

= (1.0k)( -j1591.6)

1.0k + (-j1591.6)

= 716.97-j450.47Ω

I_S(p-p) = E/Z_T

= 8 / (716.97 – j450.74)

= 9.45 < 32.14° mA

I_R(p-p) = V_R / R

= 7.6 / 1.0k

= 0.0076 A

= 0.0076<0˚

I_C(p-p) = V_C / Z_C

= 7.6<0˚ / (-j1591.6)

= 0.00478<90˚ A

R_{S measured} = 9.9Ω and V_Rs(p-p) = 0.48V

D₁ = 1 x 10ˉ⁴ D₂ = 25µs

The angle between I_R and I_S :

= 25µs x 360°

1 x 10ˉ⁴

= 90˚

I_S(p-p) = V_Rs / R_S

= 0.48V / 9.9

= 0.0483<0˚A