Series-Parallel Circuits

In reality, like in houses and buildings, connecting loads is not as easy as the ones we see in series and parallel circuits samples. In big areas like this, loads are often connected in combination of series and parallel. Such connection is called as network.

Below is an example of a simple network. Two resistors connected in parallel is connected in series to another resistor.


Example 1.

A resistor is connected in series with two other resistors which are connected in parallel. The resistance of the resistors are \(R_1=12\;\Omega,\;R_2=16\;\Omega,\;\text{and}\;R_3=4\;\Omega \) respectively. The electric potential of the circuit is \(120\;V\).

Find:

  • Resistance, \(R_p\) of the resistors in parallel
  • resistance, \(R_T \) of the circuit
  • Total current, \(I_T\) through the circuit
  • Electric potential, \(V_1, \;V_2,\;V_3\)
  • Current, \(I_1,\;I_2,\;I_3\)

Solution:

a.) To solve for the unknown quantities of the circuit, we first compute for the resistance of the resistors connected in parallel.  

\(\frac{1}{R_{2-3}}=\frac{1}{R_2}+\frac{1}{R_3}\\ \quad\;=\frac {1}{16}+\frac{1}{4}\\ \quad\;=\frac{5}{16}\)

thus, \(R_{2-3}=\frac{16}{5}\).

b.) If we let \(R_2\;\text{and}\;R_3\) as a single resistor with resistance \(R_{2-3}\), we come up with a series connection with \(R_1\). Thus, the total resistance of the circuit is given by

\(R_T=R_1+R_{2-3}\\ \quad\;=12\;\Omega\;+\;3.2\;\Omega\\ \quad\;=15.2\;\Omega\).

c.) The total current through the circuit is

\(I_T=\frac{V_0}{R_T}=\frac{90\;V}{15.2\;\Omega}=5.92\;A\).

d.) Since the connection is in series, the current all throughout the circuit is the same, hence,

\(I_T=I_1=I_{2-3}=5.92\;A\).

e.) The electric potential \(V_1\) for the first load can be calcualted as

\(V_1=I_1R_1=(5.92\;A)(12\;\Omega)=71.04\;V\).

f.) In a series circuit, the total electric potential is equal to the sum of the individual potential of each load, thus,\(V_0=V_1+V_{2-3}\). Solving for the value of \(V_{2-3}\) we have 

\(90\;V=71.04+V_{2-3}\\\)

then,

\(V_{2-3}=90\;V-71.04\;V=18.96\;V\).

g.) We are now able to get the current through \(R_2\;\text{and}\;R_3\) using \(V_{2-3}=18.96\;V\) since in a parallel circuit, the electric potential is the same across all the loads.

\(I_2=\frac{V_{2-3}}{R_2}=\frac{18.96\;V}{16\;\Omega}=1.19\;A\)             and            \(I_3=\frac{V_{2-3}}{R_3}=\frac{18.96\;V}{4\;\Omega}=4.74\;A\).

 

  • Checking that the total current in a parallel circuit is equal to the sum of the current through each load:

\(I_T=I_2+I_3=1.19\;A+4.74\;A=5.9\;A\).

  • Checking that the potential energy in a series circuit is equal to the sum of the potential in each load:

\(V_0=V_1+V_{2-3}=71.04\;V+18.96\;V=90\;V\).


Example 2.

Given:

 \(V_0=120\;V\\ R_1=25\;\Omega\\ R_2=10\;\Omega\\ R_3=15\;\Omega\\ R_4=13\;\Omega\)

 

 

Find: \(R_T,\;I_T,\;I_1,\;I_2,\;I_3,\;I_4,\;V_1,\;V_2,\;V_3,\;V_4\)

Solution:

a.) Solve first for the resistance in parallel.

\(\frac{1}{R_{3-4}}=\frac{1}{R_3}+\frac{1}{R_4}\\ \quad\quad=\frac{1}{10}+\frac{1}{15}\\ \quad\quad=\frac16\;\Omega\)

therefore, \(R_{3-4}=6\;\Omega\).

b.) The total resistance of the circuit is \(R_T=R_1+R_2+R_{3-4}=25\;\Omega\;+13\;\Omega\;+6\;\Omega=44\;\Omega\).

c.) The total current through the circuit \(I_T=\frac{V_0}{R_T}=\frac{120\;V}{44\;\Omega}=2.73\;A.\)

d.) The current through the series circuit is the same all through the circuit, \(I_T=I_1=I_2=I_{3-4}=2.73\;A.\)

e.) Solving for the electrical potential across each load:

\(V_1=I_1R_1=(2.73\;A)(25\;\Omega)=68.25\;V\)

\(V_2=I_2R_2=(2.73\;A)(13\;\Omega)=35.49\;V\)

\(V_{3-4}=(I_{3-4})(R_{3-4})=(2.73\;A)(6\;\Omega)=16.38\;V\)

Since \(V_3\;\text{and}\;V_4\) are connected in parallel, they have the same electric potential \(V_{3-4}=V_3=V_4=16.38\;V\).

f.) The current \(I_3\;\text{and}\;I_4\) is calculated as

\(I_3=\frac{V_{3-4}}{R_2}=\frac{16.38\;V}{10\;\Omega}=1.64\;A\)       and        \(I_4=\frac{V_{3-4}}{R_4}=\frac{16.38\;V}{15\;\Omega}=1.09\;A.\)