Main formulae for Part 2 Electrical principles and technology
A.c. theory
2 2 2 i 1 +i 2 +i 3 + · · · + i n I = n
1 1 T = or f = f T
2 Im or 0.637Im π 1 I = √ Im or 0.707Im 2 r.m.s maximum Form factor = Peak factor = average r.m.s
D.c. transients C−R circuit
τ = CR
Charging:
v C = V (1 − e−(t /CR) ) vr = V e−(t /CR) i = I e−(t /CR)
Discharging:
v C = v R = V e−(t /CR) i = I e−(t /CR) L τ= R
Part 2
For a sine wave: I AV =
General sinusoidal voltage: v = Vm sin(ωt ± φ)
1 X L = 2πfL XC = 2πfC V 2 2 Z = = (R + X ) I
Q=
VL V
differential voltage gain dB common mode gain −R f Vo = Inverter: A= Vi Ri Rf Vo Non-inverter: A= =1+ Vi Ri V1 V2 V3 Summing: Vo = −R f + + R1 R2 R3 1 Vi dt Vo = − Integrator: CR Rf Differential If V1 > V2 : Vo = (V1 − V2 ) − R1 If V2 > V1 :
or L C
1 R2 − 2 LC L
Vo = (V2 − V1 )
√ Star: I L = I p VL = 3V p √ Delta: VL = V p I L = 3I p √ P = 3VL I L cos φ or P = 3I p2 R p
P = VI cos φ or
Two-wattmeter method P = P1 + P2
S = VI
Q = VI sin φ
power factor = cos φ =
R3 R2 + R3
Three-phase systems
VRC L RD = L CR 2π fr L IC Q= = R Ir
Ir =
I2R
i = I (1 − e−(Rt /L) )
CMRR = 20 log10
Parallel resonance (LR–C circuit): 1 fr = 2π
v R = V (1 − e−(Rt /L) )
Operational amplifiers
1 √ 2π LC
VC 2π fr L 1 1 = = = V R 2π f r CR R fr Q= or f2 − f1 fr ( f2 − f 1) = Q
Current growth: v L = Ve−(Rt /L)
Current decay: v L = v R = V e−(Rt /L) i = I e−(Rt /L)
Single-phase circuits
Series resonance: fr =
L−R circuit
R Z
√ (P1 − P2 ) tan φ = 3 (P1 + P2 )
1+
Rf R1
Copyrighted material - Taylor & Francis Main formulae for Part 2
V1 N1 I2 = = V2 N2 I1
I0 =
2 + I2) (I M C
I M = I0 sin φ0
IC = I0 cos φ0
E2 − E1 E = 4.44 f m N Regulation = × 100% E2 2 V1 Equivalent circuit: Re = R1 + R2 V2 2 V1 Xe = X1 + X2 V2 Z e = (Re2 + X e2 ) losses Efficiency, η = 1 − input power Output power = V2 I2 cos φ2 Total loss = copper loss + iron loss Input power =output power +losses 2 N1 Resistance matching: R1 = RL N2
D.c. machines General e.m.f. E =
2 pn Z ∝ ω c
(c = 2 for wave winding, c =2 p for lap winding)
Generator: E = V + Ia Ra
VI Efficiency, η = × 100% VI + Ia2 Ra + I f V + C E = V − I a Ra
Motor:
Efficiency, η =
VI − Ia2 Ra − I f V − C VI
Torque =
× 100%
pZIa EIa = ∝ Ia 2πn πc
Three-phase induction motors f ns = s= p
n s − nr ns
× 100
fr = s f
Xr = s X2
N2 s E1 I 2 R2 Er N1 s= r Ir = = Zr P2 [R22 + (s X 2 )2 ] Efficiency, η =
Pm Pl
input – stator loss – rotor copper loss – friction and windage loss = input power
s E 12 R2 m(N2 /N1 )2 Torque, T = 2πn s R22 + (s X 2 )2 ∝
s E 12 R2 R22 + (s X 2 )2
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