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|Impedance = Voltage / Current||Z = E / I|
Similarly, when a voltage E is applied across an impedance Z, the resulting current I through
the impedance is equal to the voltage E divided by the impedance Z.
|Current = Voltage / Impedance||I = E / Z|
Similarly, when a current I is passed through an impedance Z, the resulting voltage drop V
across the impedance is equal to the current I multiplied by the impedance Z.
|Voltage = Current * Impedance||V = IZ|
Alternatively, using admittance Y which is the reciprocal of impedance Z:
|Voltage = Current / Admittance||V = I / Y|
Similarly, at any instant the algebraic sum of all the currents at any circuit node is zero:
SI = 0
Kirchhoff's Voltage Law
At any instant the sum of all the voltage sources in any closed circuit is equal to the sum of all the voltage drops in that circuit:
SE = SIZ
Similarly, at any instant the algebraic sum of all the voltages around any closed circuit is zero:
SE - SIZ = 0
Any linear voltage network which may be viewed from two terminals can be replaced by a voltage-source equivalent circuit comprising a single voltage source E and a single series impedance Z. The voltage E is the open-circuit voltage between the two terminals and the impedance Z is the impedance of the network viewed from the terminals with all voltage sources replaced by their internal impedances.
Any linear current network which may be viewed from two terminals can be replaced by a current-source equivalent circuit comprising a single current source I and a single shunt admittance Y. The current I is the short-circuit current between the two terminals and the admittance Y is the admittance of the network viewed from the terminals with all current sources replaced by their internal admittances.
The open circuit, short circuit and load conditions of the Norton model are:
Voc = I / Y
Isc = I
Vload = I / (Y + Yload)
Iload = I - VloadY
Thévenin model from Norton model
Voltage = Current / Admittance
Impedance = 1 / Admittance
E = I / Y
Z = Y -1
Norton model from Thévenin model
Current = Voltage / Impedance
Admittance = 1 / Impedance
I = E / Z
Y = Z -1
When performing network reduction for a Thévenin or Norton model, note that:
- nodes with zero voltage difference may be short-circuited with no effect on the network current distribution,
- branches carrying zero current may be open-circuited with no effect on the network voltage distribution.
In a linear network with multiple voltage sources, the current in any branch is the sum of the currents which would flow in that branch due to each voltage source acting alone with all other voltage sources replaced by their internal impedances.
If a voltage source E acting in one branch of a network causes a current I to flow in another branch of the network, then the same voltage source E acting in the second branch would cause an identical current I to flow in the first branch.
If the impedance Z of a branch in a network in which a current I flows is changed by a finite amount dZ, then the change in the currents in all other branches of the network may be calculated by inserting a voltage source of -IdZ into that branch with all other voltage sources replaced by their internal impedances.
If any number of admittances Y1, Y2, Y3, ...
meet at a common point P, and the voltages from another point N to the free ends of these admittances
are E1, E2, E3, ... then the voltage between
points P and N is:
VPN = (E1Y1 + E2Y2 + E3Y3 + ...) / (Y1 + Y2 + Y3 + ...)
VPN = SEY / SY
The short-circuit currents available between points P and N due to each of the voltages
E1, E2, E3, ... acting through the respective
admitances Y1, Y2, Y3, ... are
E1Y1, E2Y2, E3Y3,
... so the voltage between points P and N may be expressed as:
VPN = SIsc / SY
When a current I is passed through a resistance R, the resulting power P
dissipated in the resistance is equal to the square of the current I multiplied by the
P = I2R
By substitution using Ohm's Law for the corresponding voltage drop V (= IR) across the
P = V2 / R = VI = I2R
Note that power is zero for an open-circuit (zero current) and for a short-circuit (zero voltage).
When a load resistance RT is connected to a voltage source ES with series resistance RS, maximum power transfer to the load occurs when RT is equal to RS.
Under maximum power transfer conditions, the load resistance RT, load voltage
VT, load current IT and load power PT are:
RT = RS
VT = ES / 2
IT = VT / RT = ES / 2RS
PT = VT2 / RT = ES2 / 4RS
When a load conductance GT is connected to a current source IS with shunt conductance GS, maximum power transfer to the load occurs when GT is equal to GS.
Under maximum power transfer conditions, the load conductance GT, load current
IT, load voltage VT and load power PT are:
GT = GS
IT = IS / 2
VT = IT / GT = IS / 2GS
PT = IT2 / GT = IS2 / 4GS
When a load impedance ZT (comprising variable resistance RT and variable reactance XT) is connected to an alternating voltage source ES with series impedance ZS (comprising resistance RS and reactance XS), maximum power transfer to the load occurs when ZT is equal to ZS* (the complex conjugate of ZS) such that RT and RS are equal and XT and XS are equal in magnitude but of opposite sign (one inductive and the other capacitive).
When a load impedance ZT (comprising variable resistance RT and
constant reactance XT) is connected to an alternating voltage source ES
with series impedance ZS (comprising resistance RS and reactance
XS), maximum power transfer to the load occurs when RT is equal to the
magnitude of the impedance comprising ZS in series with XT:
RT = |ZS + XT| = (RS2 + (XS + XT)2)½
Note that if XT is zero, maximum power transfer occurs when RT is equal to the magnitude of ZS:
RT = |ZS| = (RS2 + XS2)½
When a load impedance ZT with variable magnitude and constant phase angle (constant power
factor) is connected to an alternating voltage source ES with series impedance
ZS, maximum power transfer to the load occurs when the magnitude of ZT
is equal to the magnitude of ZS:
(RT2 + XT2)½ = |ZT| = |ZS| = (RS2 + XS2)½
Similarly, using admittances:
YAB = YANYBN / (YAN + YBN + YCN)
YBC = YBNYCN / (YAN + YBN + YCN)
YCA = YCNYAN / (YAN + YBN + YCN)
In general terms:
Zdelta = (sum of Zstar pair products) / (opposite Zstar)
Ydelta = (adjacent Ystar pair product) / (sum of Ystar)
Similarly, using admittances:
YAN = YCA + YAB + (YCAYAB / YBC) = (YABYBC + YBCYCA + YCAYAB) / YBC
YBN = YAB + YBC + (YABYBC / YCA) = (YABYBC + YBCYCA + YCAYAB) / YCA
YCN = YBC + YCA + (YBCYCA / YAB) = (YABYBC + YBCYCA + YCAYAB) / YAB
In general terms:
Zstar = (adjacent Zdelta pair product) / (sum of Zdelta)
Ystar = (sum of Ydelta pair products) / (opposite Ydelta)
Updated 09 June 2008
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