Experimenting – Learning – Understanding Key formulae for Electrical Engineering Ohm´s law
A
I U R
U = IR
I R
V
U
Current in A Voltage in V Resistance in Ω
1V 1Ω = 1A
Resistors connected in series
I
1
U
2
U
U = U1 + U2 + U3 + ... R = R1 + R2 + R3 + ...
1
R
2
R
I =
U R
3
3
U U U U = 1 = 2 = 3 R3 R R1 R2
U
U Total voltage in V U1, U2 Voltages across individual resistors in V R Total resistance in Ω R1, R2 Resistance of individual resistors in Ω
Resistors connected in parallel
U
I I
1
R
1
I
2
I I1, I2
1 1 1 1 = + + ... + R R1 R2 R3
R
U
R
2
I
I = I1 + I2 + I3 + ... G = G1 + G2 + G3 + ...
= I •R
=
I1 • R1 = I2 • R2
3
R
3
Total current in A Current through individual resistors in A Total resistance in Ω
G Overall conductance in S G1, G2 Conductance of individual resistors in S R1, R2 Resistance of individual resistors in Ω
Kirchhoff´s first law (current law)
I
4
I
3
I
5
I
2
I
1
At any junction in an electrical circuit, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point. Σ Iin = Σ Iout
Example: I1 = 3 A; I2 = 6 A; I3 = 2 A; I4 = 1,5 A; I5 = ? A Solution: I1 + I5 = I2 + I3 + I4 I5 = I2 + I3 + I4 – I1 I5 = 6 A + 2 A + 1,5 A - 3 A I5 = 9,5 A - 3 A = 6,5 A
Kirchhoff´s second law (voltage law)
U1
U2 U5
I
U3
U4
Umlauf
In any closed current loop the directed sum of the electrical potential differences around the loop must be zero. Directed means that if you trace around the loop in a specific direction (either direction is just as valid) voltages in the same direction as you are tracing are considered positive and those in the opposite direction count as negative.
Example: U1 = 4,5 V; U2 = 1,5 V; U4 = 1 V; U5 = 3 V; U3 = ? V; Solution: U1 + U2 – U5 – U4 – U3 = 0 U3 = U1 + U2 – U4 – U5 = 4,5 V + 1,5 V – 3 V – 1 V U3 = 2 V
ΣU=0 Wheatstone’s bridge To achieve balance (I12 = 0):
1 R
R
X
R
3
N
2
RX R3 = RN R4
RX Unknown resistance in Ω RN Variable resistance in Ω R3, R4 Known bridge resistors in Ω
R
4
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