Notation | ||||||
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C E e G I i k L M N P |
capacitance voltage source instantaneous E conductance current instantaneous I coefficient inductance mutual inductance number of turns power |
[farads, F] [volts, V] [volts, V] [siemens, S] [amps, A] [amps, A] [number] [henrys, H] [henrys, H] [number] [watts, W] |
Q q R T t V v W F Y y |
charge instantaneous Q resistance time constant instantaneous time voltage drop instantaneous V energy magnetic flux magnetic linkage instantaneous Y |
[coulombs, C] [coulombs, C] [ohms, W] [seconds, s] [seconds, s] [volts, V] [volts, V] [joules, J] [webers, Wb] [webers, Wb] [webers, Wb] |
The voltages V1 and V2 which appear across the respective resistances
R1 and R2 are:
V1 = ISR1 = ESR1 / RS
= ESR1 / (R1 + R2)
V2 = ISR2 = ESR2 / RS
= ESR2 / (R1 + R2)
In general terms, for resistances R1, R2, R3, ...
connected in series:
IS = ES / RS
= ES / (R1 + R2 + R3 + ...)
Vn = ISRn = ESRn / RS
= ESRn / (R1 + R2 + R3 + ...)
Note that the highest voltage drop appears across the highest resistance.
Alternatively, when conductances G1, G2, G3, ... are
connected in parallel, the total conductance GP is:
GP = G1 + G2 + G3 + ...
where Gn = 1 / Rn
For two resistances R1 and R2 connected in parallel, the total resistance
RP is:
RP = R1R2 / (R1 + R2)
RP = product / sum
The resistance R2 to be connected in parallel with resistance R1 to give
a total resistance RP is:
R2 = R1RP / (R1 - RP)
R2 = product / difference
The currents I1 and I2 which pass through the respective resistances
R1 and R2 are:
I1 = VP / R1 = IPRP / R1
= IPR2 / (R1 + R2)
I2 = VP / R2 = IPRP / R2
= IPR1 / (R1 + R2)
In general terms, for resistances R1, R2, R3, ... (with
conductances G1, G2, G3, ...) connected in parallel:
VP = IPRP = IP / GP
= IP / (G1 + G2 + G3 + ...)
In = VP / Rn = VPGn
= IPGn / GP
= IPGn / (G1 + G2 + G3 + ...)
where Gn = 1 / Rn
Note that the highest current passes through the highest conductance (with the lowest resistance).
Alternatively, by differentiation with respect to time:
dq/dt = i = C dv/dt
Note that the rate of change of voltage has a polarity which opposes the flow of current.
The capacitance C of a circuit is equal to the charge divided by the voltage:
C = Q / V = òidt / V
Alternatively, the capacitance C of a circuit is equal to the charging current divided by the rate of
change of voltage:
C = i / dv/dt = dq/dt / dv/dt = dq/dv
For two capacitances C1 and C2 connected in series, the total capacitance
CS is:
CS = C1C2 / (C1 + C2)
CS = product / sum
The voltages V1 and V2 which appear across the respective capacitances
C1 and C2 are:
V1 = òiSdt / C1
= ESCS / C1
= ESC2 / (C1 + C2)
V2 = òiSdt / C2
= ESCS / C2
= ESC1 / (C1 + C2)
In general terms, for capacitances C1, C2, C3, ...
connected in series:
QS = òiSdt
= ESCS = ES / (1 / CS)
= ES / (1 / C1 + 1 / C2 + 1 / C3 + ...)
Vn = òiSdt / Cn
= ESCS / Cn = ES / Cn(1 / CS)
= ES / Cn(1 / C1 + 1 / C2 + 1 / C3 + ...)
Note that the highest voltage appears across the lowest capacitance.
The charges Q1 and Q2 which accumulate in the respective capacitances
C1 and C2 are:
Q1 = òi1dt
= EPC1 = QPC1 / CP
= QPC1 / (C1 + C2)
Q2 = òi2dt
= EPC2 = QPC2 / CP
= QPC2 / (C1 + C2)
In general terms, for capacitances C1, C2, C3, ...
connected in parallel:
QP = òiPdt
= EPCP = EP(C1 + C2 + C3 + ...)
Qn = òindt
= EPCn = QPCn / CP
= QPCn / (C1 + C2 + C3 + ...)
Note that the highest charge accumulates in the highest capacitance.
Alternatively, by integration with respect to time:
Y = òedt = LI
The inductance L of a circuit is equal to the induced voltage divided by the rate of change of
current:
L = e / di/dt = dy/dt / di/dt = dy/di
Alternatively, the inductance L of a circuit is equal to the magnetic linkage divided by the
current:
L = Y / I
Note that the magnetic linkage Y is equal to the product of the
number of turns N and the magnetic flux F:
Y = NF = LI
If the self induced voltages of the inductances L1 and L2 are respectively
E1s and E2s for the same rates of change of the current that produced the
mutually induced voltages E1m and E2m, then:
M = (E2m / E1s)L1
M = (E1m / E2s)L2
Combining these two equations:
M = (E1mE2m / E1sE2s)½
(L1L2)½ = kM(L1L2)½
where kM is the mutual coupling coefficient of the two inductances L1 and
L2.
If the coupling between the two inductances L1 and L2 is perfect, then the
mutual inductance M is:
M = (L1L2)½
When two coupled inductances L1 and L2 with mutual inductance M
are connected in series, the total inductance LS is:
LS = L1 + L2 ± 2M
The plus or minus sign indicates that the coupling is either additive or subtractive, depending on the
connection polarity.
Similarly, the time constant CR represents the time to change the charge on the capacitance from zero to CE at a constant charging current E / R (which produces a rate of change of voltage E / CR across the capacitance).
If a voltage E is applied to a series circuit comprising a discharged capacitance C and a
resistance R, then after time t the current i, the voltage vR across
the resistance, the voltage vC across the capacitance and the charge qC
on the capacitance are:
i = (E / R)e - t / CR
vR = iR = Ee - t / CR
vC = E - vR = E(1 - e - t / CR)
qC = CvC = CE(1 - e - t / CR)
If a capacitance C charged to voltage V is discharged through a resistance R, then after
time t the current i, the voltage vR across the resistance, the voltage vC across the capacitance and the charge qC on the capacitance are:
i = (V / R)e - t / CR
vR = iR = Ve - t / CR
vC = vR = Ve - t / CR
qC = CvC = CVe - t / CR
Inductance and resistance
The time constant of an inductance L and a resistance R is equal to L / R, and
represents the time to change the current in the inductance from zero to E / R at a constant rate of
change of current E / L (which produces an induced voltage E across the inductance).
If a voltage E is applied to a series circuit comprising an inductance L and a resistance
R, then after time t the current i, the voltage vR across the
resistance, the voltage vL across the inductance and the magnetic linkage
yL in the inductance are:
i = (E / R)(1 - e - tR / L)
vR = iR = E(1 - e - tR / L)
vL = E - vR = Ee - tR / L
yL = Li = (LE / R)(1 - e - tR / L)
If an inductance L carrying a current I is discharged through a resistance R, then after
time t the current i, the voltage vR across the resistance, the voltage vL across the inductance and the magnetic linkage
yL in the inductance are:
i = Ie - tR / L
vR = iR = IRe - tR / L
vL = vR = IRe - tR / L
yL = Li = LIe - tR / L
Rise Time and Fall Time
The rise time (or fall time) of a change is defined as the transition time between the 10% and 90% levels
of the total change, so for an exponential rise (or fall) of time constant T, the rise time (or fall
time) t10-90 is:
t10-90 = (ln0.9 - ln0.1)T » 2.2T
The half time of a change is defined as the transition time between the initial and 50% levels of the total
change, so for an exponential change of time constant T, the half time t50 is :
t50 = (ln1.0 - ln0.5)T » 0.69T
Note that for an exponential change of time constant T:
- over time interval T, a rise changes by a factor 1 - e -1
(» 0.63) of the remaining change,
- over time interval T, a fall changes by a factor e -1
(» 0.37) of the remaining change,
- after time interval 3T, less than 5% of the total change remains,
- after time interval 5T, less than 1% of the total change remains.
Similarly, the power P dissipated by a conductance G carrying a current I with a
voltage drop V is:
P = V2G = VI = I2 / G
The power P transferred by a capacitance C holding a changing voltage V with charge
Q is:
P = VI = CV(dv/dt) = Q(dv/dt) = Q(dq/dt) / C
The power P transferred by an inductance L carrying a changing current I with
magnetic linkage Y is:
P = VI = LI(di/dt) = Y(di/dt)
= Y(dy/dt) / L
Similarly, the energy W consumed over time t due to power P dissipated in a
conductance G carrying a current I with a voltage drop V is:
W = Pt = V2tG = VIt = I2t / G
The energy W stored in a capacitance C holding voltage V with charge Q is:
W = CV2 / 2 = QV / 2 = Q2 / 2C
The energy W stored in an inductance L carrying current I with magnetic linkage
Y is:
W = LI2 / 2 = YI / 2
= Y2 / 2L
The load voltage VL and load current IL for a load resistance
RL are:
VL = ILRL = EB - ILRB
= EBRL / (RB + RL)
IL = VL / RL = (EB - VL) / RB
= EB / (RB + RL)
The battery short-circuit current Isc is:
Isc = EB / RB
= EBIL / (EB - VL)
The power P dissipated by the resistance RS with voltage drop
(V - VV) carrying current IV is:
P = (V - VV)2 / RS = (V - VV)IV
= IV2RS
The power P dissipated by the resistance RP with voltage drop
VA carrying current (I - IA) is:
P = VA2 / RP = VA(I - IA)
= (I - IA)2RP
If the value of resistance R4 is unknown and the values of resistances
R3, R2 and R1 at the balance point
are known, then:
R4 = R3R2 / R1
Updated 09 June 2008 Copyright ©1998-2008 BOWest Pty Ltd |
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