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close this bookIntroduction to Electrical Engineering - Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.)
close this folder3. Electric Circuits
close this folder3.3. Branched and Unbranched Circuits
View the document3.3.1. Branched Circuits
View the document3.3.2. Unbranched Circuits
View the document3.3.3. Meshed Circuits

3.3.1. Branched 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 I1, I2 and I3 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 I1 ... I3 branching off from point A must be equal to the incoming total current I; in point B the sum of the incoming currents I1 ... I3 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 + I1 + I2 + I3

(3.2)

Since the voltage through all loads is the same, we have

I/U = I1/U + I2/U + I3/U

I/U is the conductance of the load; the total conductance or equivalent conductance, which is to be designated Gequ, is written as

Gequ = G1 + G2 + G3

(3.3)

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 n-equal conductances (n = 2, 3, 4 ...) holds

Gequ = n · G

(3.3. a)

For two consumers with the conductances G1 and G2 we have

Gequ = G1 + G2

(3.3. b)

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/Requ = 1/R1 + 1/R2 + 1/R3

(3.4.)

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 n-equal resistances (n = 2, 3, 4 ...) holds

Requ = R/n

(3.4. a)

For two loads with the resistances R1 and R2 the following holds:

1/Requ = 1/R1 + 1/R2 = (R1 + R2)/(R1 · R2)

Requ = (R1 · R2)/(R1 + R2)

(3.4. b)

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

I1/I2 = G1/G2

(3.5 a)

or

I1/I2 = G1/Gequ

(3.5 a)

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.)

I1/I2 = R1/R2

(3.6.a)

I1/I2 = Requ/R1

(3.6.b)

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 G1 = 12.2 mS and G2 = 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:

Gequ
Requ
I
I1; I2

Solution:

Circuit:


Fig. 3.7. Circuit with two loads connected, in parallel

According to equation (3.5.b):

Gequ = G1 + G2
Gequ = 12.2 mS + 8.7 mS
Gequ = 20.9 mS;

the equivalent conductance is greater than the greatest individual conductance

According to equation (2.3.):

Requ = 1/Gequ
Requ = 1/0.0209 S = 48 1/S
Requ = 48 W

or according to equation (5.4.b) where for

R1 = 1/G1 = 1/0.0122 S = 82 W and for
R2 = 1/G2 = 1/0087 S = 115 W

we have to write

Requ = (R1 · R2)/(R1 + R2)
Requ = (82 W · 115 W)/(82 W + 115 W) = 9430 W/197W = 47.9 W
Requ » 48 W

the equivalent resistance is smaller than the smallest individual resistance

According to equation (3.1.b):

I = U/Requ
I = 24 V/48 W
I = 0.5 A = 500 mA

I1 = U/R1 = 24 V/82 W
I1 = 0.292 A = 292 mA

the greater partial current flows through the smaller individual resistance

I2 = U/R2 = 24 V/115 W
I2 = 0.208 A = 208 mA

I1 and I2 can also be found with the help of the current divider rule, i.e. equations (3.5. and 3.6.), namely,

I1/I = G1/Gequ

or

I1/I = Requ/R1

I1 = I · G1/Gequ


I1 = I · (Requ/R1)

I1 = 500 mA · (12.2/20.9)


I1 = 500 mA · (48/82)

I1 = 292 mA


I1 = 292 mA

The calculation for I2 has to be performed analogously.

Check with the help of equation (3.2.):

I = I1 + I2
I = 292 mA + 208 mA
I = 500 mA

the total current is equal to the sum of all partial currents