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

Meshed Circuits, also known as interconnected circuits, are such circuits where loads are connected in parallel and other loads in series. In addition, the various “branches can contain voltage sources. Computations of these circuits are possible with the help of the two Kirchhoff’s laws. Here, we confine ourselves to the loaded voltage divider as the simplest case which is of great importance to practical electrical engineering.
By means of a voltage divider, one can draw from an available voltage any desired smaller partial voltage which is used to supply a certain load R. Through the latter, an adequate current will then flow. Besides a voltage division, a current division is also involved; Fig. 3.12. shows these conditions.
Fig. 3.12. Loaded voltage divider
R = R_{equ1}
Circuits of this type where resistances are connected in parallel and other resistances in series are called seriesparallel connections. According to Fig. 3.12., the following considerations should be made:
Through the load R_{V} a current passes which is part of the total current. As a consequence, the current flowing through R_{1} only has the intensity of I_{1} = I  I_{V}, i.e. it is smaller than in the case of a nonconnected load. A lower current causes a smaller voltage drop according to Ohm’s law. This fact is described by the following equation.
U_{1}/U = R_{equ1}/R_{equ}
R_{equ} = (R_{1} · R_{V})/(R_{1} + R_{V}) is smaller than the smaller resistance of R_{1} and R_{V}
R_{equ} = R_{equ1} + R_{2}
Example 3.6.
From a voltage source of 24 V, a voltage of 6 V is to be supplied t a load R_{V} = 30 W via a voltage divider. The total current is 0.5 A. Calculate the partial resistances R_{1} and R_{2}
Given:
U = 24 V
I = 0.5 A
U_{1} = 6V
R_{V} = 30 W
To be found:
R_{1} and R_{2}
Solution:
R_{2} = U_{2}/I
R_{2} = 18 V/0.5 A
R_{2} = 36 W
U_{2} = U  U_{1} = 24 V  6 V =18 VR_{equ1} = U_{1}/I
R_{equ1} = 6 V/0.5 A = 12 W
1/R_{equ1} = 1/R_{1} +1/R_{V} inverted for R_{1}
1/R_{1} = 1/R_{equ1}  1/R_{V} = (R_{V}  R_{equ1})/(RV · R_{equ1})
R_{1} = (R_{V} · R_{equ1})/(R_{V}  R_{equ1})
R_{1} = (30 W · 12 W)/(30 W  12 W) = 360/18 W
R_{1} = 20 W
Verification by means of equation (3.10.):
U_{1} = U · (R_{equ1}/R_{equ})
U_{1}= 25 V · (12 W/(12 W + 36 W)) = 24 V · 12/48
U_{1} = 6 V_{}
If the load had not been connected, then for the selected divider resistances the voltage dropping via R_{1}, would have been greater and, according to equation (3.9.b) would be
U_{1} = U · (R_{1}/R_{equ})
U_{1}= 24 V · (20/36) = 40/3 V
U_{1} = 13.3 V_{}
According to their circuit structures, a distinction is made between branched, unbranched and meshed circuits. In branched circuits all loads are connected in parallel. Their equivalent conductance is equal to the sum of the individual conductances, hence, always greater than the greatest individual conductance. The equivalent resistance, however, is always smaller than the smallest individual resistance.
G_{equ} = G_{1} + G_{2} + G_{3} + ...
1/R_{equ} = 1/R_{1} + 1/R_{2} + 1/R_{3} + ...
The voltage across all loads is equal. At each junction point,
Fig. 3.13. Circuit for example 3.6.
the sum of the currents flowing toward, the point is equal to the sum of those flowing away from it.
I = I_{1} + I_{2} + I_{3} + ...
The division of the voltage by the loads is effected in the same ratio as that of their conductances.
I_{1}/I = G_{1}/G_{equ}; I_{2}/I = G_{2}/G_{equ}; I_{3}/I = G_{3}/G_{equ} etc.
In unbranched circuits, all loads are connected in series. Their equivalent resistance is equal to the sum of the individual resistances, hence always greater than the greatest individual resistance.
R_{equ} = R_{1} + R_{2} + R_{3} + ...
The current is equal through all loads. The total voltage is equal to the sum of all individual voltage drops.
U = U_{1} + U_{2} + U_{3} + ...
Voltage division by the loads is effected in the same ratio as that of their resistances.
U_{1}/U = R_{1}/R_{equ}; U_{2}/U = R_{2}/R_{equ}; U_{3}/U = R_{3}/R_{equ} etc.
In meshed circuits, some of the loads are connected in parallel and some of them in series. Such a circuit is called seriesparallel connection. For computing or designing a seriesparallel connection, the laws at the bottom of both branched and unbranched circuits have to be taken into account. A seriesparallel connection of particular importance to electrical engineering is the loaded voltage divider.
Questions and problems:
1. The following loads are connected in parallel to a voltage of 110 V: 1 soldering iron having 200 W, 1 incandescent lamp having 120 W, 1 heater having 24 W. Calculate the partial currents and, with the help of the 1st Kirchhoff’s law, the total current. Check the result according to Ohm’s law after having determined the equivalent resistance of the three loads connected in parallel.2. By means of a certain ammeter, a current of maximum I_{i} = 10Ma can be measured. Assume the resistance of the instrument is R_{i} = 10 W. In order to measure a higher current, for example, a current of I = 100 mA, a resistor of a suitable resistance value must be connected in parallel through which (that is to say, past the instrument) a current of 90 mA (in the case of fullscale deflection) or, in general, of I  I_{i} can flow. Calculate the value of the parallel resistance for the conditions given above.
3. In an electronic device, three resistors are connected in series to a voltage of 12 V: R_{1} = 360 W, R_{2} = 240 W, R_{3} = 120Q. Calculate the equivalent resistance, the current and the partial voltage dropping via R_{1} to R_{2}.
4. By means of a certain voltmeter, a maximum voltage of U_{i} = 15 mV can only be measured. Assume the resistance of the instrument is R_{i} = 5 W. In order to be in a position to measure a higher voltage, for example, a voltage of U = 3 V, a resistor of a suitable resistance value must be connected in series with the voltmeter. Via the latter, the excessive voltage of 2.985 V (in the case of fullscale deflection) or, in general, of U  U_{i} can drop. Calculate the value of the series resistor for the conditions specified above.
5. Two resistors R_{1} = 18 kW and R_{2} = 36 kW are connected in series and to a voltage of U = 110 V. A load of R_{V} = 10 kW is connected in parallel to R_{1}. Calculate the voltage at R_{V}, the total current and the partial currents.