Parallel Circuits

In a Parallel Circuit the electric charges reach each load moving or passing through different path. Each load has its corresponding path from the source. The electrons may flow from load to load one at a time, depending on which path it will take first and then go back to the source. Meaning, the charges have a choice to which path it will take.

Given:

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

electric current = \(I_0,\;I_1,\;I_2,\;\text{and}\;I_3\)

resistance = \(R_p,\;R_1,\;R_2,\;\text{and}\;R_3\)

The electric potential of each load in a parallel circuit is the same as the total electric potential of the circuit. Mathematically expressed as

\(V_0=V_1=V_2=V_3\)                                                          (1) 

The total current in the circuit is equal to the sum of the current flowing through individual loads of the circuit,

\(I_0=I_1+I_2+I_3\)                                                             (2)

Using Ohm's Law, \(I=\frac VR\), electric current can be solved as

\(I_0=\frac {V_0}{R_p}\quad\quad I_1=\frac {V_1}{R_1} \quad\quad I_2=\frac {V_2}{R_2}\quad \text{and}\quad I_3=\frac {V_3}{R_3}\)               (3)

Substituting (3) to (2), we have

\(\frac {V_0}{R_p}=\frac {V_1}{R_1}+\frac {V_2}{R_2}+\frac {V_3}{R_3}\)                                                           (4)

But, using (1), the total resistance in a parallel circuit is given by

\(\frac {1}{R_p}=\frac {1}{R_1}+\frac {1}{R_2}+\frac {1}{R_3}\)                                                             (5)


Example 1.

Two resistors with resistance \(R_1=3\;\Omega\) and \(R_2=7\;\Omega\) are connected to a source of \(60\;V\).

 

 

Find: \(V_1,\;V_2,\;R_p,\;I_0,\;I_1,\;I_2\)

Solution:

a.) Since electric potential is the same all throughout the circuit, and \(V_0=60\;V\), then

\(V_0=V_1=V_2=V_3=60\;V\).

b.) The total resistance of the load in the parallel circuit is

\(\frac {1}{R_p}=\frac {1}{R_1}+\frac {1}{R_2}\\ \quad\;=\frac {1}{3}+\frac {1}{7}\\ \quad\;=\frac{10}{21}\)

then \(R_p=\frac{21}{10} \;\Omega=2.1\;\Omega\).

c.) The current through each resistor

\(I_1=\frac{V_1}{R_1}=\frac{60\;V}{3\;\Omega}=20\;A\)

\(I_2=\frac{V_2}{R_2}=\frac{60\;V}{7\;\Omega}=8.57\;A\)

d.) In solving for the total electric current, we can use either of the two solutions:

    d.1.) Using (2)

 \(I_0=I_1+I_2=20\;A+8.57\;A=28.57\;A\)

    d.2.) or using (3)

\(I_0=\frac{V_0}{R_p}=\frac{60\;V}{2.1\;\Omega}=28.57\;A\)


Example 2.

Given: \(R_1=2\;\Omega,\;R_3=6\;\Omega,\;R_p=\frac{12}{11}\;\Omega \;\text{and}\;V_1=20\;V \)

 

 

Find: \(R_2,\; V_0,\;V_2,\;V_3,\;I_0,\;I_1,\;I_2,\;I_3\)

Solution:

a.) To find \(R_2\), use (5).

\(\frac {1}{R_p}=\frac {1}{R_1}+\frac {1}{R_2}+\frac {1}{R_3}\)

\(\frac {1}{R_p}= \begin{pmatrix} \frac {1}{R_1}+\frac {1}{R_3} \end{pmatrix} +\frac {1}{R_2}\)

\(\frac {1}{\frac{12}{11}}= \begin{pmatrix} \frac {1}{2\;\Omega}+\frac {1}{6\;\Omega} \end{pmatrix} +\frac {1}{R_2}\)

\(\frac{11}{12}= \begin{pmatrix} \frac {4}{6} \end{pmatrix} +\frac {1}{R_2}\)

\(\frac{11}{12}-\begin{pmatrix} \frac {4}{6} \end{pmatrix}=\frac {1}{R_2}=\frac{3}{12}\)

thus, \(R_2=4\;\Omega.\)

b.) From (1) and \(V_1=20\;V\), the electric potential across the circuit is

\(V_0=V_1=V_2=V_3=20\;V\).

c.) The current through the circuit

\(I_1=\frac{20\;V}{2\;\Omega}=10\;A\)

\(I_2=\frac{20\;V}{4\;\Omega}=5\;A\)

\(I_3=\frac{20\;V}{6\;\Omega}=\frac{10}{3}\;A=3.33\;A\)

d.) The total current along the circuit is

\(I_0=I_1+I_2+I_3=10\;A+5\;A+3.33\;A=18.33\;A\).