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 V_{1} and V_{2} which appear across the respective resistances
R_{1} and R_{2} are:
V_{1} = I_{S}R_{1} = E_{S}R_{1} / R_{S}
= E_{S}R_{1} / (R_{1} + R_{2})
V_{2} = I_{S}R_{2} = E_{S}R_{2} / R_{S}
= E_{S}R_{2} / (R_{1} + R_{2})
In general terms, for resistances R_{1}, R_{2}, R_{3}, ...
connected in series:
I_{S} = E_{S} / R_{S}
= E_{S} / (R_{1} + R_{2} + R_{3} + ...)
V_{n} = I_{S}R_{n} = E_{S}R_{n} / R_{S}
= E_{S}R_{n} / (R_{1} + R_{2} + R_{3} + ...)
Note that the highest voltage drop appears across the highest resistance.
Alternatively, when conductances G_{1}, G_{2}, G_{3}, ... are
connected in parallel, the total conductance G_{P} is:
G_{P} = G_{1} + G_{2} + G_{3} + ...
where G_{n} = 1 / R_{n}
For two resistances R_{1} and R_{2} connected in parallel, the total resistance
R_{P} is:
R_{P} = R_{1}R_{2} / (R_{1} + R_{2})
R_{P} = product / sum
The resistance R_{2} to be connected in parallel with resistance R_{1} to give
a total resistance R_{P} is:
R_{2} = R_{1}R_{P} / (R_{1} - R_{P})
R_{2} = product / difference
The currents I_{1} and I_{2} which pass through the respective resistances
R_{1} and R_{2} are:
I_{1} = V_{P} / R_{1} = I_{P}R_{P} / R_{1}
= I_{P}R_{2} / (R_{1} + R_{2})
I_{2} = V_{P} / R_{2} = I_{P}R_{P} / R_{2}
= I_{P}R_{1} / (R_{1} + R_{2})
In general terms, for resistances R_{1}, R_{2}, R_{3}, ... (with
conductances G_{1}, G_{2}, G_{3}, ...) connected in parallel:
V_{P} = I_{P}R_{P} = I_{P} / G_{P}
= I_{P} / (G_{1} + G_{2} + G_{3} + ...)
I_{n} = V_{P} / R_{n} = V_{P}G_{n}
= I_{P}G_{n} / G_{P}
= I_{P}G_{n} / (G_{1} + G_{2} + G_{3} + ...)
where G_{n} = 1 / R_{n}
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 C_{1} and C_{2} connected in series, the total capacitance
C_{S} is:
C_{S} = C_{1}C_{2} / (C_{1} + C_{2})
C_{S} = product / sum
The voltages V_{1} and V_{2} which appear across the respective capacitances
C_{1} and C_{2} are:
V_{1} = òi_{S}dt / C_{1}
= E_{S}C_{S} / C_{1}
= E_{S}C_{2} / (C_{1} + C_{2})
V_{2} = òi_{S}dt / C_{2}
= E_{S}C_{S} / C_{2}
= E_{S}C_{1} / (C_{1} + C_{2})
In general terms, for capacitances C_{1}, C_{2}, C_{3}, ...
connected in series:
Q_{S} = òi_{S}dt
= E_{S}C_{S} = E_{S} / (1 / C_{S})
= E_{S} / (1 / C_{1} + 1 / C_{2} + 1 / C_{3} + ...)
V_{n} = òi_{S}dt / C_{n}
= E_{S}C_{S} / C_{n} = E_{S} / C_{n}(1 / C_{S})
= E_{S} / C_{n}(1 / C_{1} + 1 / C_{2} + 1 / C_{3} + ...)
Note that the highest voltage appears across the lowest capacitance.
The charges Q_{1} and Q_{2} which accumulate in the respective capacitances
C_{1} and C_{2} are:
Q_{1} = òi_{1}dt
= E_{P}C_{1} = Q_{P}C_{1} / C_{P}
= Q_{P}C_{1} / (C_{1} + C_{2})
Q_{2} = òi_{2}dt
= E_{P}C_{2} = Q_{P}C_{2} / C_{P}
= Q_{P}C_{2} / (C_{1} + C_{2})
In general terms, for capacitances C_{1}, C_{2}, C_{3}, ...
connected in parallel:
Q_{P} = òi_{P}dt
= E_{P}C_{P} = E_{P}(C_{1} + C_{2} + C_{3} + ...)
Q_{n} = òi_{n}dt
= E_{P}C_{n} = Q_{P}C_{n} / C_{P}
= Q_{P}C_{n} / (C_{1} + C_{2} + C_{3} + ...)
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 L_{1} and L_{2} are respectively
E_{1s} and E_{2s} for the same rates of change of the current that produced the
mutually induced voltages E_{1m} and E_{2m}, then:
M = (E_{2m} / E_{1s})L_{1}
M = (E_{1m} / E_{2s})L_{2}
Combining these two equations:
M = (E_{1m}E_{2m} / E_{1s}E_{2s})^{½}
(L_{1}L_{2})^{½} = k_{M}(L_{1}L_{2})^{½}
where k_{M} is the mutual coupling coefficient of the two inductances L_{1} and
L_{2}.
If the coupling between the two inductances L_{1} and L_{2} is perfect, then the
mutual inductance M is:
M = (L_{1}L_{2})^{½}
When two coupled inductances L_{1} and L_{2} with mutual inductance M
are connected in series, the total inductance L_{S} is:
L_{S} = L_{1} + L_{2} ± 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 v_{R} across
the resistance, the voltage v_{C} across the capacitance and the charge q_{C}
on the capacitance are:
i = (E / R)e^{ - t / CR}
v_{R} = iR = Ee^{ - t / CR}
v_{C} = E - v_{R} = E(1 - e^{ - t / CR})
q_{C} = Cv_{C} = 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 v_{R} across the resistance, the voltage v_{C} across the capacitance and the charge q_{C} on the capacitance are:
i = (V / R)e^{ - t / CR}
v_{R} = iR = Ve^{ - t / CR}
v_{C} = v_{R} = Ve^{ - t / CR}
q_{C} = Cv_{C} = 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 v_{R} across the
resistance, the voltage v_{L} across the inductance and the magnetic linkage
y_{L} in the inductance are:
i = (E / R)(1 - e^{ - tR / L})
v_{R} = iR = E(1 - e^{ - tR / L})
v_{L} = E - v_{R} = Ee^{ - tR / L}
y_{L} = 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 v_{R} across the resistance, the voltage v_{L} across the inductance and the magnetic linkage
y_{L} in the inductance are:
i = Ie^{ - tR / L}
v_{R} = iR = IRe^{ - tR / L}
v_{L} = v_{R} = IRe^{ - tR / L}
y_{L} = 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) t_{10-90} is:
t_{10-90} = (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 t_{50} is :
t_{50} = (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 = V^{2}G = VI = I^{2} / 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 = V^{2}tG = VIt = I^{2}t / G
The energy W stored in a capacitance C holding voltage V with charge Q is:
W = CV^{2} / 2 = QV / 2 = Q^{2} / 2C
The energy W stored in an inductance L carrying current I with magnetic linkage
Y is:
W = LI^{2} / 2 = YI / 2
= Y^{2} / 2L
The load voltage V_{L} and load current I_{L} for a load resistance
R_{L} are:
V_{L} = I_{L}R_{L} = E_{B} - I_{L}R_{B}
= E_{B}R_{L} / (R_{B} + R_{L})
I_{L} = V_{L} / R_{L} = (E_{B} - V_{L}) / R_{B}
= E_{B} / (R_{B} + R_{L})
The battery short-circuit current I_{sc} is:
I_{sc} = E_{B} / R_{B}
= E_{B}I_{L} / (E_{B} - V_{L})
The power P dissipated by the resistance R_{S} with voltage drop
(V - V_{V}) carrying current I_{V} is:
P = (V - V_{V})^{2} / R_{S} = (V - V_{V})I_{V}
= I_{V}^{2}R_{S}
The power P dissipated by the resistance R_{P} with voltage drop
V_{A} carrying current (I - I_{A}) is:
P = V_{A}^{2} / R_{P} = V_{A}(I - I_{A})
= (I - I_{A})^{2}R_{P}
If the value of resistance R_{4} is unknown and the values of resistances
R_{3}, R_{2} and R_{1} at the balance point
are known, then:
R_{4} = R_{3}R_{2} / R_{1}
Updated 09 June 2008 Copyright ©1998-2008 BOWest Pty Ltd |
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