Introduction to Electrical Engineering  Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.) 
3. Electric Circuits 
3.3. Branched and Unbranched Circuits 

When connecting several consumers, also known as loads, in parallel in a circuit, then the current can flow at the same time through several consumers. Fig. 3.6. shows such an arrangement.
Fig. 3.6. Branched circuit
The current I driven by the voltage source is divided into three partial currents I_{1}, I_{2} and I_{3} at branching point A. These currents pass through the three loads, join in point B and return as total current I to the voltage source. Since no charge carriers are lost, the sum of the partial currents I_{1 }... I_{3 }branching off from point A must be equal to the incoming total current I; in point B the sum of the incoming currents I_{1 }... I_{3} must be equal to the total current I returning to the source.
In general, the statement formulated by Robert Kirchhoff (German physicist 1824  1887) and known as the 1st Kirchhoff’s law or junction point theorem holds;
In each junction point, the sum of the currents flowing toward the point is equal to the sum of those flowing away from it.
For the example shown in Fig. 3.6. thus, we have
I + I_{1} + I_{2} + I_{3}
Since the voltage through all loads is the same, we have
I/U = I_{1}/U + I_{2}/U + I_{3}/U
I/U is the conductance of the load; the total conductance or equivalent conductance, which is to be designated G_{equ}, is written as
G_{equ} = G_{1} + G_{2} + G_{3}
In a parallel connection of consumers, the equivalent conductance is equal to the sum of the individual conductances; it is always greater than the greatest individual conductance.
For nequal conductances (n = 2, 3, 4 ...) holds
G_{equ} = n · G
For two consumers with the conductances G_{1} and G_{2} we have
G_{equ} = G_{1} + G_{2}
In practice, resistances are more frequently used for calculating than conductances. In accordance with the general relation R = 1/G, the following is derived from equation (3.3)
1/R_{equ} = 1/R_{1} + 1/R_{2} + 1/R_{3}
In a parallel connection of loads, the reciprocal value of the equivalent resistance is equal to the reciprocal values of the individual resistances; the equivalent resistance is always smaller than the smallest individual resistance.
For nequal resistances (n = 2, 3, 4 ...) holds
R_{equ} = R/n
For two loads with the resistances R_{1} and R_{2} the following holds:
1/R_{equ} = 1/R_{1} + 1/R_{2} = (R_{1} + R_{2})/(R_{1 }·_{ }R_{2})
R_{equ} = (R_{1} · R_{2})/(R_{1} + R_{2})
The ratio of the partial currents is dependent on the ratio of the partial resistances. It is obvious that the smaller current flows through the larger partial resistance and vice versa. This relation which is known as the current divider rule for two loads connected in parallel is written as
I_{1}/I_{2} = G_{1}/G_{2}
or
I_{1}/I_{2} = G_{1}/G_{equ}
In a current divider, the ratio of the partial currents is like that of the partial conductances.
Expressed in terms of resistances, we obtain from equation (3.5.)
I_{1}/I_{2} = R_{1}/R_{2}
I_{1}/I_{2} = R_{equ}/R_{1}
In a current divider, the ratio of the partial currents is inverse to that of the partial resistances.
Example 3.4
Two loads with the conductances G_{1} = 12.2 mS and G_{2} = 8.7 mS are connected in parallel while a voltage of 24 V is applied to them. Draw a sketch of the circuit. Calculate the equivalent conductance and the equivalent resistance as well as the total current and the partial currents.
Given:
G1 = 12.2 mS
G2 = 8.7 mS
U = 23 V
To be found:
G_{equ}
R_{equ}
I
I_{1}; I_{2}
Solution:
Circuit:
Fig. 3.7. Circuit with two loads
connected, in parallel
According to equation (3.5.b):
G_{equ} = G_{1} + G_{2}
G_{equ} = 12.2 mS + 8.7 mS
G_{equ} = 20.9 mS;
the equivalent conductance is greater than the greatest individual conductance
According to equation (2.3.):
R_{equ} = 1/G_{equ}
R_{equ} = 1/0.0209 S = 48 1/S
R_{equ} = 48 W
or according to equation (5.4.b) where for
R_{1} = 1/G_{1} = 1/0.0122 S = 82 W and for
R_{2} = 1/G_{2} = 1/0087 S = 115 W
we have to write
R_{equ} = (R_{1} · R_{2})/(R_{1} + R_{2})
R_{equ} = (82 W · 115 W)/(82 W + 115 W) = 9430 W/197W = 47.9 W
R_{equ} » 48 W
the equivalent resistance is smaller than the smallest individual resistance
According to equation (3.1.b):
I = U/R_{equ}
I = 24 V/48 W
I = 0.5 A = 500 mA
I_{1} = U/R_{1} = 24 V/82 W
I_{1} = 0.292 A = 292 mA
the greater partial current flows through the smaller individual resistance
I_{2} = U/R_{2} = 24 V/115 W
I_{2} = 0.208 A = 208 mA
I_{1} and I_{2} can also be found with the help of the current divider rule, i.e. equations (3.5. and 3.6.), namely,
I_{1}/I = G_{1}/G_{equ} 
or 
I_{1}/I = R_{equ}/R_{1} 
I_{1} = I · G_{1}/G_{equ} 

I_{1} = I · (R_{equ}/R_{1}) 
I_{1} = 500 mA · (12.2/20.9)  
I_{1} = 500 mA · (48/82) 
I_{1} = 292 mA  
I_{1} = 292 mA 
The calculation for I_{2} has to be performed analogously.
Check with the help of equation (3.2.):
I = I_{1} + I_{2}
I = 292 mA + 208 mA
I = 500 mA
the total current is equal to the sum of all partial currents