  Introduction to Electrical Engineering - Basic vocational knowledge (Institut f³r Berufliche Entwicklung, 213 p.)  3. Electric Circuits  3.3. Branched and Unbranched Circuits  3.3.1. Branched Circuits 3.3.2. Unbranched Circuits 3.3.3. Meshed Circuits

3.3.2. 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 = E1 + E2

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 en-tire energy to the circuit, the statement in the form of the 2nd Kirchhoff’s law or the mesh-network 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 = E1 + E2 = U = U1 + U2 + U3

(3.7)

As the current through all loads is the same, we have

U/I = U1/I + U2/I +U3/I

U/I, however, is the resistance of the load; the total resistance or equivalent resistance will be called Requ and obtained in the form of

Requ = R1 + R2 + R3

(3.8.)

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

Requ = n · R holds.

(3.8.a)

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:

U1/U2 = R1/R2

(3.9.a)

or

U1/U = R1/Requ

(3.9.b)

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 R1 = 18 W and R2 = 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:

R1 = 18 W
R2 = 72 W
U = 60 V

To be found:

I
U1; U2

Solution:

Circuit:

According to equation (3.8.):

Requ = R1 +R2
Requ = 18 W + 72 W
Requ = 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/Requ
I = 60 V/90 W
I = 0.0667 A = 667 mA

According to equation (3.1.a):

U1 = R1 · I
U1 = 18 W · 0.667 A
U1 = 12 V;

the smaller partial voltage drops via the smaller individual resistance

U2 = R2 · I
U2 = 72 W · 0.667 A
U2 = 48 V

U1 and U2 can also be calculated according to the voltage divider rule, equation (3.9.b), namely,

 U1/U = R1/Requ or U2/U = R2/Requ U1 = U · (R1/Requ) U2 = U · (R2/Requ) U1 = 60 V · 18/90 U2 = 60 V · (72/90) U1 = 12 U2 = 48

Verification by means of equation (3.7.):

U = U1 +U2
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 = I1 + I2 U = I1 + U2 Gequ = G1 + G2 Requ = R1 + R2 1/Requ = 1/R1 + 1/R2 1/Gequ = 1/G1 + 1/G2 ® Requ = (R1 · R2)/(R1 + R2) ® Gequ = (G1 · G2)/(G1 + G2) for G1 = G2 = G = 1/R is for R1 = R2 = R = 1/G is Gequ = 2 · G, since n = 2 Requ = 2 · R, since n = n or Requ = R/2 or Gequ = G/2 I1/I2 = G1/G2 = R2/R1 U1/U2 = R1/R2 = G2/G1 I1/I = G1/Gequ = G1/(G1 + G2) U1/U = R1/Requ = R1/(R1 + R2) I1/I = Requ/R1 = R2/(R1 + R2) U1/U = Gequ/G1 = G2/(G1 + G2)