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 load in one circuit in such a manner that the current passes through all loads one after the other, then this circuit is called series connection or series circuit.
Fig. 3.8. shows such an arrangement.
Fig. 3.8. Unbranched circuit
The two voltage sources connected in series produce a total voltage of
E = E_{1} + E_{2}
The current driven by the total voltage has the same intensity at any point in the circuit and causes an adequate voltage drop at any load. Since the charge carriers give their entire energy to the circuit, the statement in the form of the 2nd Kirchhoff’s law or the meshnetwork theorem holds:
In each unbranched circuit, the sum of all voltage drops is equal to the total voltage.
For the example shown in Fig. 3.8., thus we have
E = E_{1} + E_{2} = U = U_{1} + U_{2} + U_{3}
As the current through all loads is the same, we have
U/I = U_{1}/I + U_{2}/I +U_{3}/I
U/I, however, is the resistance of the load; the total resistance or equivalent resistance will be called R_{equ} and^{ }obtained in the form of
R_{equ }= R_{1} + R_{2} + R_{3}
In a series connection of loads, the equivalent resistance is equal to the sum of the individual resistances; it is always greater than the greatest individual resistance.
For nequal resistances (n = 2, 3, 4, ...)
R_{equ }= n · R holds.
If conductances have to be calculated, the required equations can easily be derived from the relation G = 1/R. In practice, calculations of conductances are required very seldom for unbranched circuits.
The ratio of the partial voltages is dependent on the ratio of the partial resistances. It is obvious that the greater voltage drop occurs via the greater partial resistance and vice versa. This relation called voltage divider rule is expressed for two loads connected in aeries in the following way:
U_{1}/U_{2} = R_{1}/R_{2}
or
U_{1}/U = R_{1}/R_{equ}
In a voltage divider, the ratio of the partial voltages is the same as that of the partial resistances.
Example 3.5.
Two loads with the resistances R_{1} = 18 W and R_{2} = 72 W are connected in series and a voltage of 60 V is applied to them. Draw a sketch of this circuit. Calculate the equivalent resistance, the current and the partial voltages.
Given:
R_{1} = 18 W
R_{2} = 72 W
U = 60 V
To be found:
I
U_{1}; U_{2}
Solution:
Circuit:
According to equation (3.8.):
R_{equ} = R_{1} +R_{2}
R_{equ} = 18 W + 72 W
R_{equ} = 90 W
the equivalent resistance is greater than the greatest individual resistance
Fig. 3.9. Circuit with two loads
connected in series
According to equation (3.1.b):
I = U/R_{equ}
I = 60 V/90 W
I = 0.0667 A = 667 mA
According to equation (3.1.a):
U_{1} = R_{1} · I
U_{1} = 18 W · 0.667 A
U_{1} = 12 V;
the smaller partial voltage drops via the smaller individual resistance
U_{2} = R_{2} · I
U_{2} = 72 W · 0.667 A
U_{2} = 48 V
U_{1} and U_{2} can also be calculated according to the voltage divider rule, equation (3.9.b), namely,
U_{1}/U = R_{1}/R_{equ} 
or 
U_{2}/U = R_{2}/R_{equ} 
U_{1} = U · (R_{1}/R_{equ}) 

U_{2} = U · (R_{2}/R_{equ}) 
U_{1} = 60 V · 18/90  
U_{2} = 60 V · (72/90) 
U_{1} = 12 

U_{2} = 48 
Verification by means of equation (3.7.):
U = U_{1} +U_{2}
U = 12 V + 48 V
U = 60 V;
the sum of all voltage drops is equal to the total voltage.
In the following Table 3.2., an unbranched circuit is compared with a branched circuit. In each of the two cases, two loads are included. It is obvious that the formulas of one circuit have the same structure as those of the other circuit. The only difference is that, as compared to the branched circuit, for the unbranched circuit U is put instead of I, I instead of U, R instead of G and G instead of R.
Table 3.2. Relations in Branched and Unbranched Circuits Having two Loads Each
Branched Circuit 
Unbranched Circuit 


U is equal across all loads 
I is equal through all loads 
 
I = I_{1} + I_{2} 
U = I_{1} + U_{2} 
 
G_{equ} = G_{1} + G_{2} 
R_{equ} = R_{1} + R_{2} 
 
1/R_{equ} = 1/R_{1} + 1/R_{2} 
1/G_{equ} = 1/G_{1} + 1/G_{2} 
 
® R_{equ} = (R_{1} · R_{2})/(R_{1} + R_{2}) 
® G_{equ} = (G_{1} · G_{2})/(G_{1} + G_{2}) 
 
for G_{1} = G_{2} = G = 1/R is 
for R_{1} = R_{2} = R = 1/G is 
 
G_{equ} = 2 · G, since n = 2 
R_{equ} = 2 · R, since n = n 
 
or R_{equ} = R/2 
or G_{equ} = G/2 
 
I_{1}/I_{2} = G_{1}/G_{2} = R_{2}/R_{1} 
U_{1}/U_{2} = R_{1}/R_{2} = G_{2}/G_{1} 


I_{1}/I = G_{1}/G_{equ} = G_{1}/(G_{1} + G_{2}) 
U_{1}/U = R_{1}/R_{equ} = R_{1}/(R_{1} + R_{2}) 
 
I_{1}/I = R_{equ}/R_{1} = R_{2}/(R_{1} + R_{2}) 
U_{1}/U = G_{equ}/G_{1} = G_{2}/(G_{1} + G_{2}) 