|
E I V Y Z |
Voltage source Current Voltage drop Admittance Impedance |
(Volts) (Amps) (Volts) (Siemens) (Ohms) |
| Voltage = Current * Impedance | V = IZ |
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 |
Kirchoff's Voltage Law
The sum of all the voltage sources in any closed circuit is equal to the sum of all the
voltage drops in that circuit.
Similarly, the sum of all the voltages around any closed circuit is zero.
Any 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 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.
|
Voltage = Current / Admittance Impedance = 1 / Admittance |
E = I / Y Z = Y -1 |
Norton from Thévenin
|
Current = Voltage / Impedance Admittance = 1 / Impedance |
I = E / Z Y = Z -1 |
In a 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 small 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 + ...)
Similarly, using admittances:
YAB = YANYBN / (YAN + YBN + YCN)
YBC = YBNYCN / (YAN + YBN + YCN)
YCA = YCNYAN / (YAN + YBN + YCN)
In general terms:
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:
Ystar = (sum of Ydelta pair products)
/ (opposite Ydelta)