sambungan jawapan ca2 lab2

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

D₁ = 5 X 10ˉ⁴s D₂ = 25µs

The angle between I_R and I_C :

= 25µ x 360°

5X10ˉ⁴

= 18°

I_C(p-p) = V_Rs / R_S

= 0.48 / 9.9

= 0.0483A

By using Ohm’s law and R_measured = 0.987kΩ :

I_R(p-p) = V_R / R → V_R = E = 8V

= 8 / 0.987k

= 8.11mA

I_S(p-p) = I_R(p-p) + I_C(p-p)

= 8.11m + 0.0483

= 0.05641 A

By using the value of E_(p-p) = 8V and I_S(p-p) = 0.05641 A :

Z_T = E / I_S

= 8 / 0.05641

= 141.82 Ω

X_C = -j

wC

= -j

2πf(C)

= -j

2π(10k)(0.01µ)

= -j1591.6 Ω

Calculation for phasor diagram

I_C = 0.0483 < 90°mA

I_S

θ

Θ_S

I_R = 8.11mA

By using theorem pithagoras :

I_S =

= ^*

= 8.11 A

Θ_S = tan^-1 (0.0483 / 8.11m)

= 80.47°

Therefore

I_S = 8.11 < 80.47°mA

Based on phasor diagram :

Θ_S = 80.47°

Θ = 33.79°

Θ_C = 90°