Easy question:

1 B -I leads V by 90°
-V lags I by 90°
So the correct answer is (B)
2 a) C 1
XC = 2πFc -frequency doubled since speed of rotation doubled
b) C
a) F doubled
XC halved (C)
NAB2πF 2
b) I = 1 =F
2πFC
I ∝ F2
I increase 4 times (C)
3 A C1 + C2 = 6 + 3 = 9 nF
18 nF // 9 nF = 6nF
4 B VR leads VC by 90°

5 B IC = IR
So they are in the same phase
6 C (R – L - C)
V −V X −X 3R− 2R R
Tanθ = LV C = L R C = R = R = 1
R
Tan θ = 1
θ = 45°
7 A 16 x 32
Ct = 16+32 = 10.66
10.66 x 8 32
Ct = 8+10.66 = 7 µF
8 B Ohmic resistance circuit
V, I in the same phase
9 C 1
XC = 2πFc
XC
Slope = 1 = XC x C
C
2000 1
= = 500
1000 x 103
1 1
= 2πF
500
500 = 2πF
F = 79.57Hz
10 C (C - circuit)
NAB2πF
Imax = 2πFc
Imax ∝ F2
11 D PF → x 10-12F
1+1=2
1+1=2
2 // 2
2x2
Ct = 2+2 = 1Pf
12 C R only
13 B (R – L - C)
X −X 3R−2R R
Tanθ = L R C = R = R = 1
θ = 45°
14 E Vt leads It
When VL > VC
XL > XC (inductive reactance)
15 A - Coil stores energy as magnetic field
- Capacitor stores energy as electric energy

Medium questions:
1 C Cap + Dc = open key
So the value of (I) vanished at complete charge
2 C (R – C - circuit)
X X 1 X
Tanθ = Rc → Tan 30 = Rc → = RC
√3
R
XC =
√3

Z = √XC2 + R2
R 2 2√3
Z = √( ) + R2 = R
√3 3
Z 2√3
=
R 3
3 D XCtotal = 50Ω
1
XCtotal = 2πFC
total
1
50 = 500 → Ct = 2 x 10−5 F × 106 = 20µF
2π x x Ct
3.14

C +C = 2C
2
2C // C = 3C
2
C = 20
3
C = 30 µF
4 A (C - circuit)
V 500
XCtotal = I = 2 = 250Ω
1 1
XCtotal = 2πFc → 250 = 100
total 2π x x Ctotal
π
Ctotal = 2 x 10-5 → x 106 = 20μF
30 x 15
= 10 → 10 + C = 20
30+15
C = 10µF
5 B Case (1):
R = XC
 Z1 = √Xc2 + R2
= √R2 + R2
Z1 = R√2
Case 2:
1
XC ∝ F → Frequency doubled so XC halfed
R is halfed also
Z2 = √Xc2 + R2
1 1
= √(2 R)2 + (2 R)2
1
Z2 = R
√2
Z1 R√2
= 1 =2
Z2 R
√2
Z1
Z2 = 2
6 A (L - circuit)
V leads I by 90°
So I lags V by 90°
7 B C → 1.5 R so I increases
B → XL decreases and Z also decreases so I increases
A → C increases but XC decreases and also Z decreases so I increases
I increases in all circuits
8 1) C 1) A: (coil)
2) B XL = 2πFL high frequency → XL high → I won’t pass
So bulb (A) won’t light
1
B: XC = 2πFC high frequency → XC low → I will pass
So bulb (B) lights
2) Blub (A):
XL = 2πFL frequency low → XL low → I ↑
so bulb (A) lights
bulb (B):
1
XC = 2πFC frequency low → XC high → I (won’t pass)
So blub (B) won’t light
9 B Cap + Dc = open key
So bulb (B) won’t light, only bulb (A) lights
10 B Veff 1
Ieff = = 1
Xc
2πFC
Ieff = wc
Ieff ∝ w
11 B Q = CV
In series, Q constant
So the charges on the plates of the capacitor equal each other
12 D Q = CV
Q constant
 Cy > C x 2C > C
1
Vy < V x Vy = 2Vx
So figure (D) is the answer
13 D 1
Xctotal = 2πFC
total
1
50 = 500 → Ctotal = 2 x 10−5 F x 106 = 20µF
2π x x Ctotal
π

C + C = 2C
2 2
2C // C → 3C → 3C = 20
C = 30µF
14 B V 500
XCtotal = total = = 250Ω
Itotal 2
1 1
250 = 2π FC → 250 = 100
2π x x Ctotal
π
Ctotal = 2 x 10-5F
x 106 = 20µF
30 // 15 = 10µF
10 + C = 20
C = 10µF
15 C (R – L – C circuit)
XL = 3R Xc = 2R
XL − XC 3R−2R R
Tanθ = R = R = R = 1
θ = 45°
16 D XC = 2XL = 2R
XC > XL
X −X 2R−R R
Tan θ = c R L = R = R = 1
θ = 45°
V lags VR by 45°
17 C Induction characteristics (XL > XC) at point (C)
18 D Veff = Vmax x sin45 = 100 x sin45 = 50√2
Ieff = Imax x sin45 = 10 x sin45 = 5√2
P = Veff x Ieff
P = 50√2 x 5√2 = 500W
19 D (R – L - circuit)
XC > XL
X −X 2R−R R
Tan θ = c R L = R = R = 1
θ = 45°
So V lags VR by 45°
20 B XL = 2πFL high frequency ↑ so XL high ↑→ so reading (A1)
decreases
1
XC = 2πFC high frequency ↑ so XC low ↓ so reading (A2) increases
21 A At resonance: Z = R
 As Fe core removed:
2
Z = √(XL2 − XC ) + R2
So Z increases, I decreases
22 C I leads V by 45°
So this circuit connected to a resistor and a capacitor
23 A (R – C circuit)
VR, I in the same phase
24 B 1
At the same frequency XC = 2πFC
XC2 > XC1
C2 < C1
25 D 1st quarter → charged
4th quarter → discharged
26 C C2 + C3 = 4 + 2 = 6µF
4x6
4µF // 6µF = 4+6 = 2.4µF
27 B Key open: (R – C circuit)
V 200
Z = I = 0.2 = 1000Ω

1000 = √R2 + XC2

1000 = √5002 + XC2

XC = 866.025Ω
Key closed:
Cdoubled → 2C
1
XC = 2πFc Cdoubled
XC halfed = 433.012Ω
Z = √R2 + XC2

= √5002 + 433.0122
= 661.4378Ω
200
I = 661.4378 = 0.3A
28 B 1 1
XC = 2πFc → XC = wc
XC
Slope = 1 = Xc x W
w
1
Slope = C
Sx > S y
Cx < C y
29 A Xc1 F
= 2 x
C2
C2 = 2C // 2C = C
Xc2 F1 C1
C
C1 = 4
2F C
= xC
F
4
 8
=2x4=1
30 1) C 1) XC = 3R
2) B
Z = √R2 + XC2 = √R2 + 3R2 = √10R2 = √10R
V 12
2) R = I = 2 = 6Ω
31 B (L – C circuit)
V = √(VC − VL )2
Vt = VC - VL
V = VC – V
VC = 2V
32 A E = I2R
Capacitor and coil stores energy so if you want to increase the rate of energy
then you must increase R
33 A 1
Fr = 2π√LC
C ↑ 4 times
Frequency decrease to half
But L remains constant
34 B XC = 2XL
XC > XL (capacitive)
So the phase angle is negative
35 D Imax → resonance
1 1
2πFL = 2πFC → 2πFL = 2πFπ
1
2Πfl = 2π2 x F
4π3F2L = 1
1
L = 4π3 F2 → there is no correct answer (D)
36 D Fr1 L xC
= √L2 x C2 L2 = 4 L1
Fr2 1 1
C
C2 = 91
C
F 14L x 1
= √ L x C9
F2 1 1

F 1
= √4 x
F2 9
F 2
=3
F2
3F
F2 = 2
37 B 1 1
Fr = 2π →W=
√LC √LC
1
W= = 2500rad/sec.
√8 x 10−3 x 20 x 10−6
220
I= = 5A
44
38
39
40 B L
= 25 L = 25C
C
1 1
Fr = 2π → 1592 =
√LC 2π √25c2
-5
C = 2 x 10 F
x 106 = 20µF
41 D XC = 2XL = 2R
XC > XL
Vt lags VR
X −X 2R−R R
Tanθ = C R L = R = R = 1
θ = 45°
Vt lags VR by 45°
42 B Fr1 L xC
= √L 2 x C2
Fr2 1 1

10 2L x 2C 10
= √ = √4
F2 LC F2
10
=2
F2
F2 = 5KHz
43 C X L = XC resonance
So the circuit act as ohmic
44 A b:
V, I in same phase
X L = XC
c:
V leads I
X L > XC (inductive)
a:
I leads V
XC > XL (capacitive)
45 C XL − XC 60−80
Tan θ = R = 20 = −1
θ = −45°

Hard questions:

1 A I = 2 x 10-3
Q = cv → 12 x 10−6 = 3 x 10−6 x V
V = 4V
Vab = 4 + (2 x 10-3 x 4 x 103) – 15 = 3V
2 A C2 + C3 = 3 + 3 = 6µF
6µF // 3µF = 2
Qt = C x V
 = 2 x 4 = 8µC
Q 8 x 10−6
C1 → V1 = C = = 2.7V
3 x 10−6
V2 = V3 4 – 2.7 = 1.3V
3 D I2 = 1A I1 = 3A
_(3 x 6) + (2 x 1) = VC
18 + 2 = VC
VC = 20V
Q=CxV
Q = 4 x 10-6 x 20 = 8 x 10-5C
4 B 8µF // 8µF = 4µF
4µF + 1µF = 5µF
10
2 // 5 = µF
7
10 24
+ 2 = 7 µF
7
5 B Vt = Vcoil + Vy → algebraic sum
(Pure)
Induction coil of zero resistance
6 C Vt = √(VL − VC )2
Vt = VL – VC (L - C circuit)
So the element must be pure capacitor
7 0.5
XL = 2πFL = 2π x 50 x π = 50Ω
1 1
XC = 2πFC = 1 = 20
2π x 50 x
2000π

Z = √(XL − XC )2 + R2 = √(50 − 20)2 + 402 = 50

a) XL = 50 (ture)
b) False
c) Z = 50 (false)
V 100
d) I = Z = 50 = 2A (true)
e) V = I x XL V = 2 x 50 = 100V (true)
f) V = I x R = 2 x 40 = 80V (false)
g) V = I x XC = 2 x 20 = 40V (false)
h) Vt = √VL2 + VR2 = √1002 + 802 = 128.4 V (true)

i) Vt = √402 + 802 = 89.44V (false)

j) Vt = √(100 − 40)2 + 802 = 100V (true)

k) True
XL 50
l) Tanθ = = 40 θ = 51.34° (false)
R
−XC −20
m) Tanθ = = A = 26.56 (false)
R 40
XL−XC 50−20 3
n) Tanθ = = =4 θ = 36.9° (true)
R 40
o) XC = XL
 So we must add +30 not 15 (false)
1
8 C Z = √(XL − XC )2 + R2 = ((XL − XC )2 + R2 ) 2

9 - C (resonance)
- C 1 1
- Fr = 2π√LC → −4
= 50.32Hz
2π√0.1 x 10
- Remains constant
XL = XC
10 C 2Πf = 1000
F = 159.15Hz XL= 2πFL = 1000 x 2 = 2000Ω
V 100
It = Rt = 500 = 0.2A
1 1 1
Xc = 2πFC = WC = = 2000Ω
1000 x 0.5 x 10−6
X L = XC (resource)
11 D 1
Fr = 2π√LC
1
a) Fr = = 159.15 (not correct)
2π√1 x 10−6 x 1
1
b) Fr = = 15.91 (not correct)
2π√10 x 10−6 x 10
1
c) Fr = = 79.57 (not correct)
2π√2 x 10−6 x 2
1
d) Fr = 7 7
= 500 (correct)
2π√ x 10−6 x
22 22
12 (3) V L = VC → XL = XC
13 C -3 // 3 // 3 → 1µF
3 + 1 = 4µF
-4 + 2 = 6µF
3 //6 = 2µF
-4 + 2 = 6µF
6 // 3 = 2µF
Cx2
-C //2 → C+2 = 1
C = 2µF

Problems:
1 - L – circuit:
Vmax NAB2πF
I= = 2πFL
XL
I independent on F
- C – circuit:
Vmax NAB2πF
I= X = 1
C
2πFC
I ∝ F2
2 - L – circuit:
NAB2πF
I = 2πFL
I independent on F
 I remains constant
- C – circuit:
I ∝ F2
3F → I = 9 times
- R- circuit:
NAB2πF
I= → I ∝F
R
3F → I increase 3 times
3 X
Tan θ = RC
XC 1 XC
Tan30 = → =
R √3 R
√3 x 1
R =XC√3 → R = → R2πFc = √3
2πFc
√3
F = R2πc (R – C circuit)
4 Tan60 =
XC
→ √3 =
XC
→ √3R = XC
R R
1
√3R = 2πFC → 2πFc x √3R = 1
1
(2πFCR)2 = ( )2
√3
(2πFCR)2 = 0.33
5 At resource → Z = R
1) a) Z increase, I decrease
P = I2R → so the power decrease
b) Z = √XL2 + R2 so Z increase
2) a) brightness remains constant (resource)
b) XL = 2πFL→ F doubled so XL doubled also
1
XC = 2πFC → F doubled so XC halfed also
1
XL − XC 2XL − XC
2
tan θ = =
R R
1
c) XL = 2 Xc
1x1
2πFL = 4πFC 8π2F2LC = 1
1
F2 = 8π2 LC
1 1
F= π = 2π√2LC
√8 √LC
d) XL > XC, so it is inductive
e) canceled
6 1 7 7 7
XCtotal = 2πFC Ctotal = 22 + 22 = 11
total
1 7 7 7
= 7 // 11 = 22µF
2π x 50 x x 10−6 11
22
= 10004.02Ω
Vt 10
It =X = 10004.02 = 1 x 10−3 A
Ctotal
7 Z = √(XL − XC )2 + R2
 Z= √(87.96 − 8)2 + 602
Z= 99.97 ≅ 100Ω
XL = 2πFL
XL = 2π x 50 x 0.28
XL = 87.96Ω
V 220
It = Z = 100 = 2.2A
8 Resource → XL = XC = 250Ω
1 1
a) XC = 2πFC 250 = 1000
2π x xC
44
C = 2.8 x 10−5 F
X 106 = 28.011µF
b) Z = R = 100 VL = VC
V 200
I= = =2A
R 100
VL = I x XL = 2 x 250 = 500V
9 It↑ → Xc ↓ → C ↑ (connection parallel)
200 300 500
Ctotal = 11 + 11 = 11 = 45.45µF
1
XCtotal = = 25Ω
2π x 140 x 45.45 x 10−6
V 200
I=X = = 8A
Ctotal 25
10 R.L.C (resonance) XL = XC
1 1
XC = 2πFC = = 31.83Ω
2π x 50 x 100 x 10−6
XL = 31.83Ω
V 100
I = R = 25 = 4A
11 a) Before:
Vt V
Z = √(XL − XC )2 + R2 It1 = =X
Z C
Z1 = XL – XC
= 2 Xc – XC
Z = XC
After: (XL = 2XC)
XL ∝ F → frequency doubled → XL doubles → 2XL
1 1
XC ∝ F → frequency doubled → XC halfed → 2XC
1 1
Z2 = XL – XC = 2XL - XC = 2 x 2 Xc - Xc
2 2
1
= 4 Xc - 2 Xc = 3.5 Xc
V V
It2 = Z = 3.5 Xc
Z1 XC 1 2
= = =7
Z2 3.5 XC 3.5
I1 V 3.5XC
b) = x
I2 XC V
I1 7
= 3.5 = 2
I2
11 12
a) R = = 6Ω
2
Vt 12
b) Z = = 1.2 = 10Ω
It

10 = √R2 + XL2

10 = √62 + XL2
XL = 8Ω
c) Resonance → XL = XC
12 1 1
a) XC = 2πFC = = 331.57Ω
2π x 60 x 8 x 10−6
V 150
It = X = 331.57 = 0.45A
C
b) C – circuit → I ∝ F2
Frequency doubled → I2 increase 4 times (4 x 0.45 = 1.8A)
13 1 1
XC = 2πFC = −6 = 1894.7Ω
2π x 60 x 1.4 x 10
V 120
It = X = 1894.7 = 0.063A
C
14 Veff = Vmax x sin45
= 170 x sin45 = 120.208V
V
Ieff = Xeff
C
120.208
0.75 = → XC = 160.27Ω
XC
1
160.27 = 2π x 60 x C → C = 1.7 x 10-5F
X 106 = 16.54µF
-3
15 C = 1 x 10
1
θ = 2πft = 2 x 180 x 60 x 720 = 30°
1
XC = 2πFC Z = √R2 + XC2
1
= = 2.65Ω Z = √(4.58)2 + (2.65)2 = 5.29Ω
2π x 60 x 1 x 10−3
(R -C circuit)
Xc
Tan30 = R
2.65
Tan30 = R = 4.58Ω
R
V 120
It = Z = 5.29 = 22.67A
16 Resonance
1 1
Fr = 2π√LC = −6
= 568.41Hz
2π√16 x 10 x 4.9 x 10−3
17 Canceled
18 L = 0.03 x 10-3 C = 0.005 x 10-6
1 1
Fr = 2π√LC = −3 −6
= 410936.296Hz
2π √0.03 x 10 x 0.005 x 10
19 (Resonance) L = 9 x 10-3 Fr = 980 x 103
1
Fr = 2π√LC
 1
980 x 103 = C = 2.93 x 10-12F
2π x √9 x 10−3 x C
x 1012 = 2.93PF
20 C1 = 40µF Fr1 = 750kHz L2 = 5L1 C2 = C1 + 32 = 72
F1 L2 x C2 750 5L x 72
= √L → = √ L 1x 40
F2 1 x C1 F2 1
750
=3 F2 = 250KHz
F2
21 C1 = 18µF F1 = 2 x 104Hz F2 = 3 x 104Hz L1 = L2
4
F1 L xC 2 x 10 C
= √L2 x C2 → 4 = √ 182
F2 1 1 3 x 10
C2 = 8µF
22
Z = √XL2 + R2

Z = √52 + 122 = 13Ω

XL = 2πFL
7
XL = 2 x π x 50 x 440
XL = 4.997 ≅ 5Ω
23 XL = 2πFL = 2 x π x 60 x 0.5 = 188.49Ω
Z = √XL2 + R2 = √188.492 + 402 = 192.69Ω
(R -L circuit)
X 188.49
Tanθ = RL = 40
θ = 78.01°
24 a) F = 0 → XL = 0
V 6
I = R = 6 = 1A
7
b) XL = 2πFL = 2 x π x 50 x 275 = 7.99 ≅ 8Ω

Z = √XL2 + R2 = √62 + 82 = 10Ω

V 6
I = Z = 10 = 0.6A
25 (R – L circuit)
21
XL = 2πFL = 2 x π x 50 x 220 = 29.98 ≅ 30Ω

a) Z = √XL2 + R2 = √302 + 402 = 50Ω

80
b) It = 50 = 1.6A
VL = I x XL = 1.6 x 50 = 80V
VR = I x R = 1.6 x 40 = 64V
It is not correct to add these voltages algebraically because
Vt = √VL2 + VR2 and this is vector sum not algebraic sum
26 XL = 2πFL = 2 x π x 60 x 0.4 = 150.79Ω
 Z = √XL2 + R2 = √150.792 + 602 = 162.29Ω
V 90
It= = 162.29 = 0.55A
Z

a) VR = I x R = 0.55 x 60 = 33V
b) VL = I x XL = 0.55 x 150.79 = 82.93V
27 V 50
It = R = 100 = 0.5A
Vt 130
It = → 0.5 = → Z = 260Ω
Z Z

260 = √(XL )2 + R2 → 260 = √(XL )2 + 1002

XL = 240Ω
150
a) XL = 2πFL → 240 = 2π x π x L
L = 0.8H
X 240
b) Tanθ = RL → tan θ = 100
θ = 67.38°
28 1 1
(R – C circuit) XC = 2πFC =
2π x 50 x 6 x 10−6

Z = √XC2 + R2 XC = 530.51Ω

Z = √530.512 + 103.442
Z = 540.5Ω
V 216
a) It = t = = 0.399 ≅ 0.4A
Z 540.5
−XC 530.51
b) Tanθ = R = − 103.44
θ = −78.96°
29 P = VI 60 = 100 x I
I = 0.6A (a)
V 200
I = Z → 0.6 = Z → Z = 333.33Ω
(R – C circuit)
100 2
Z = √XC2 + R2 → 333.33 = √(XC )2 + ( 0.6 )
XC = 288.67Ω
1 1
XC = 2πFC → 288.67 = 2π x 50 x C C = 1.1 x 10-5F (C)
Vcap = I x XC = 0.6 x 288.67 = 173.202V (b)
−X −288.67
d) tanθ = R C = 166.66
θ = −60°
30 P = VI
1
25 = 100 x I I = 4A
1
XC = 2πFC
1
= 100 = 300Ω
2π x 50 x x 10−6
3π
 V2 1002
P= → 25 = → R = 400Ω
R R
Z = √3002 + 400 = 500Ω 2
V 200
It = Zt = 500 = 0.4A
1
So the lamp melts (0.4 > 4)
31 Tan60 =
XC
→ √3 =
XC
→ √3R = XC
R R
1
√3R = 2πFC → 2πFC x √3R = 1
1
(2πFCR)2 = ( )2
√3
(2πFCR)2 = 0.33
32 V 200
Z = max = = 50Ω
Imax 4
- I leads V by 50°
θ = 50°
So the elements: resistor + capacitor
X X
Tan50 = RC → 1.19 = RC → XC = 1.19R

Z = √XC2 + R2
50 = √(1.19R)2 + R2
50 = 1.55R
R = 32.16Ω
XC = 1.19R
= 1.19 x 32.16 = 38.27Ω
33 Ct = 30// 15 = 10µF
30// 60 = 20 µF
10 + 10 = 20µF
20 + 20 = 40µF
1 1
XCtotal = 2πFC = = 79.57Ω
total 2π x 50 x 40 x 10−6
Vt 400
It = X = 79.57 = 5.02A
Ctotal
34 Vt = 22V
1 1 1 1 11
= + + =
Ct 1 2 3 6
6
Ct = 11µF → 10−6 F
6
Qt = C xV = 22 x x 10−6 = 1.2 x 10−5 C
11
Q 1.2 x 10−5
V1 = C = = 12V
1 x 10−6
Q 1.2 x 10−5
V2 = C = = 6V
2 x 10−6
Q 1.2 x 10−5
V3 = = = 4V
C 3 x 10−6
35 Vt = 120V
Ct = 2 + 4 = 6µF
=6 // 3 = 2 µF
 Qt = C x V = 120 x 2 x 10-6 = 2.4 x 10-4C
Q 2.4 x 10−4
X- V = C = = 80V
3 x 10−6
Plate Y, Z → V = 120 – 80 = 40V
QY = CY x VY = 2 x 10-6 x 40 = 8 x 10-5C
QZ = CZ x VZ = 4 x 10-6 x 40 = 1.6 x 10-4C
36 C = 2 x 10-6F Vmax = 100 T = 4 x 10-3S
Veff = 100 x sin45 = 50√2
1
F= −3 = 250Hz
4 x 10
1 1
XC = 2πFC = = 318.3Ω
2π x 250 x 2 x 10−6
V 50√2
It = X t = = 0.22A
C 318.3
37 a) I2 = 0 I1 = I3
15 15
It = 5+7 = 12 = 1.25A (I1 = I3)
Cap + Dc = open key
b) V = 1.25 x 5 = 6.25V
Vcap + 6 = 6.25
Vcap = 0.25V
Q = C x V = 2 x 10-6 x 0.25 = 5 x 10-7C
38 a) I1 = I4 (cap + Dc = open key)
I1 + I2 = I3 (1)
- 4I2 – 8 = 0
I2 = 2A
- 3I1 + 9 – 8 = 0
1
I1 = − 3A
−1
I4 = A
3
1
− 3 + 2 =I3
5
I3 = 3
Cap + Dc = open key I5 = 0
1
b) 3 x 3 = VCap + 7
Vcap = -6V
Q = C x V = 5 x 10-6 x 6 = 3 x 10-5C
39 XL = XC (resonance)
VL = VC Z=R
V 100
I = R = 50 = 2A
V1 = I x R = 2 x 50 = 100V
V2 = I x XL = 2 x 25 = 50V
V3 = I x Xc = 2 x 25 = 50V
V4 = 50 – 50 = 0V
40
a) Z = √(XL − XC )2 + R2 = √(20 − 16)2 + 32 = 5Ω
 V 20
b) It = Z = 5 = 4A
c) V1 = I x R = 4 x 3 = 12V
V2 = I x XL = 4 x 20 = 80V
V3 = I x XC = 4 x 16 = 64V
V4 = V2 – V3 = 80 – 64 = 16V