
calculus da abeh da~ ahahah tnggl seminggu agey.. wan da balek lol... awal neyr dier balek... ;(

VRs(p-p) = 0.48V RS measured = 9.9Ω
D1 = 5 X 10ˉ4s D2 = 25µs
The angle between IR and IC :
= 25µ x 360°
5X10ˉ4
= 18°
IC(p-p) = VRs / RS
= 0.48 / 9.9
= 0.0483A
By using Ohm’s law and Rmeasured = 0.987kΩ :
IR(p-p) = VR / R → VR = E = 8V
= 8 / 0.987k
= 8.11mA
IS(p-p) = IR(p-p) + IC(p-p)
= 8.11m + 0.0483
= 0.05641 A
By using the value of E(p-p) = 8V and IS(p-p) = 0.05641 A :
ZT = E / IS
= 8 / 0.05641
= 141.82 Ω
XC = -j
wC
= -j
2πf(C)
= -j
2π(10k)(0.01µ)
= -j1591.6 Ω
Calculation for phasor diagram
IC = 0.0483 < 90°mA
IS
θ
ΘS
IR = 8.11mA
By using theorem pithagoras :
IS =
=
= 8.11 A
ΘS = tan-1 (0.0483 / 8.11m)
= 80.47°
Therefore
IS = 8.11 < 80.47°mA
Based on phasor diagram :
ΘS = 80.47°
Θ = 33.79°
ΘC = 90°
CALCULATION
By using capacitor value = 0.01µF and measured resistance = 0.987kΩ :
Zc = -j
wC
= -j1591.6Ω
ZT = R//Zc
= (1.0k)( -j1591.6)
1.0k + (-j1591.6)
= 716.97-j450.47Ω
IS(p-p) = E/ZT
= 8 / (716.97 – j450.74)
= 9.45 < 32.14° mA
IR(p-p) = VR / R
= 7.6 / 1.0k
= 0.0076 A
= 0.0076<0˚
IC(p-p) = VC / ZC
= 7.6<0˚ / (-j1591.6)
= 0.00478<90˚ A
RS measured = 9.9Ω and VRs(p-p) = 0.48V
D1 = 1 x 10ˉ4 D2 = 25µs
The angle between IR and IS :
= 25µs x 360°
1 x 10ˉ4
= 90˚
IS(p-p) = VRs / RS
= 0.48V / 9.9
= 0.0483<0˚A
lawatan ke neutral batik kat cengal lempong... 















neyh arr gambar terbaru deecrime....