Series Circuits

Series Circuit is a type of connection wherein from the source, the electron flows in a single path going to the loads before going back to the source. This means that if there are at least 3 loads connected to the source, the electric charges will pass through each of the loads in the circuit following the same path before going back to the source.

To analyze an electric circuit, the following electrical quantities must be considered:

1. the electric potential, V or the source of energy in the circuit
2. the electric current, I that flows all throughout the circuit
3. the total resistance and individual resistance of the load

Let us analyze the sample circuit below and find out the different formulas and equations we will use in solving for the unknown quantities in a series circuit.

Given:

electric potential = $$V_0,\;V_1,\;V_2,\;\text{and}\;V_3$$

resistance = $$R_s,\;R_1,\;R_2\;\text{and}\;R_3$$

current through the circuit = $$I_0,\;I_1,\;I_2,\;\text{and}\;I_3$$

In a series circuit, the sum of the electric potential used by every load is equal to the total electric potential (source) of the circuit, that is

$$V_0=V_1+V_2+V_3$$                                                          (1)

Using the relationship $$V=IR,$$ the electric potential for the corresponding individual loads and the total electric potential can be expressed as

$$V_0=I_0R_s,\quad V_1=I_1R_1,\quad V_2=I_2R_2$$                          (2)

Substituting the values of V in (2) to (1), the equation will become

$$I_0R_s=I_1R_1+I_2R_2+I_3R_3$$                                         (3)

But, knowing that the electric current that flows all throughout a series circuit is the same,

$$I_0=I_1=I_2=I_3$$                                                          (4)

then (3) can be expressed as

$$R_s=R_1+R_2+R_3$$                                                      (5)

where $$R_s$$ denotes the total resistance of the series ciruit which is equal to the sum of the resistance of each load in the ciruit.

Example 1.

Three resistors $$R_1=10\;\Omega,\;R_2=15\;\Omega,\;\text{and}\;R_3=20\;\Omega$$ in series are connected to a source with electric potential of $$90\;V$$.

Find: $$R_s,\;I_0,\;I_1,\;I_2,\;I_3,\;V_1,\;V_2,\;V_3$$

Solution:

a.) The total resistance of a series circuit is equal to the sum of the resistances of each resistor in the circuit, using (5)

$$R_s=R_1+R_2+R_3\\ \quad\;=10\;\Omega+15\;\Omega+20\;\Omega\\ \quad\;=45\;\Omega.$$

b.) Applying the relationship $$V_0=I_0 R_s$$ in (2), the total current on the circuit is

$$I_0=\frac{V_0}{R_s}=\frac{90\;V}{45\;\Omega}=2\;A.$$

c.) Using (4), the current though the circuit is

$$I_0=I_1=I_2=I_3=2\;A$$.

d.) To get the values of the electric potential, use (2), thus

$$V_1=I_1R_1=(2\:A)(10\;\Omega)=20\;V$$

$$V_2=I_2R_2=(2\:A)(15\;\Omega)=30\;V$$

$$V_3=I_3R_3=(2\:A)(20\;\Omega)=40\;V$$

e.) Solving (1), the total electric potential of the circuit is

$$V_0=V_1+V_2+V_3\\ \quad=20\;V+30\;V+40\;V\\ \quad=90\;V.$$

Example 2.

The total resistance of the resistors in a series circuit is $$R_s=50\;\Omega$$ where $$R_1=8\;\Omega,\;R_2=12\;\Omega\;\text{and}\;R_4=6\;\Omega.$$ The current that flows through $$R_1$$ is $$I_1=2.4\;A$$. Find: $$R_3,\;I_0,\;I_2,\;I_3,\;I_4,\;V_0,\;V_1,\;V_2,\;V_3\;V_4$$.

Solution:

a.) Using (4) and $$I_1=2.4\;A$$,

$$I_0=I_1=I_2=I_3=I_4=2.4 \;A$$.

b.) $$R_3$$ can be solved using (5),

$$R_s=R_1+R_2+R_3+R_4\\ 50\;\Omega=(R_1+R_2+R_4)+R_3\\ 50\;\Omega=(8+12+6)+R_3\\ 50\;\Omega=(26)+R_3\\$$

then,

$$R_3=50\;\Omega - 26\;\Omega\\ \quad\;=24\;\Omega.$$

c.) We use (2) to calculate the electric potential on each load.

$$V_1=I_1R_1=(2.4\:A)(8\;\Omega)=19.2\;V$$

$$V_2=I_2R_2=(2.4\:A)(12\;\Omega)=28.8\;V$$

$$V_3=I_3R_3=(2.4\:A)(24\;\Omega)=57.6\;V$$

$$V_4=I_4R_4=(2.4\:A)(6\;\Omega)=14.4\;V$$.

e.) To compute the total electric potential, we can have two solutions.

e-1.) Using (1)

$$V_0=V_1+V_2+V_3+V_4\\ \quad\,=19.2\;V+28.8\;V+57.6\;V+14.4\;V\\ \quad\,=120\;V.$$

e-2.) Using (2)

$$V_0=I_0R_s=(2.4\;A)(50\;\Omega)=120\;V$$.