Published by Patrick Mutisya · 8 days ago
In any circuit, charge cannot accumulate at a junction. Therefore, the algebraic sum of currents at a junction is zero. This is often written as
\$\sum I{\text{in}} = \sum I{\text{out}}\$
For a simple parallel circuit with two branches, the rule becomes
\$I{\text{total}} = I1 + I_2\$
When components are connected end‑to‑end, the same current flows through each, and the total potential difference supplied by the source is divided among them. The rule is
\$V{\text{total}} = V1 + V2 + V3 + \dots\$
For a series chain of three resistors, the voltage across each resistor can be found from Ohm’s law \$V = IR\$ once the current is known.
All branches of a parallel network share the same two nodes, so the potential difference across each branch is identical:
\$V{\text{parallel}} = V{\text{branch 1}} = V_{\text{branch 2}} = \dots\$
This is a direct consequence of the definition of voltage as the energy change per unit charge between two points.
| Quantity | Symbol | Formula | When to Use |
|---|---|---|---|
| Current (junction rule) | \$I\$ | \$I{\text{total}} = \displaystyle\sum I{\text{branch}}\$ | Parallel circuits |
| Voltage (series rule) | \$V\$ | \$V{\text{total}} = \displaystyle\sum V{\text{component}}\$ | Series circuits |
| Voltage (parallel rule) | \$V\$ | \$V{\text{parallel}} = V{\text{any branch}}\$ | Parallel circuits |
| Resistance (parallel) | \$R_{\text{eq}}\$ | \$\frac{1}{R{\text{eq}}}= \frac{1}{R1}+ \frac{1}{R_2}+ \dots\$ | Finding total resistance of parallel network |
| Resistance (series) | \$R_{\text{eq}}\$ | \$R{\text{eq}} = R1 + R_2 + \dots\$ | Finding total resistance of series network |
Consider the circuit below (suggested diagram in the figure). A 12 V battery supplies a network that consists of a 4 Ω resistor \$R1\$ in series with a parallel combination of \$R2 = 6\ \Omega\$ and \$R_3 = 12\ \Omega\$.
Solution
Parallel equivalent:
\$\frac{1}{R_{23}} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12}= \frac{3}{12}\$
\$R_{23}= \frac{12}{3}=4\ \Omega\$
Total resistance:
\$R{\text{total}} = R1 + R_{23}=4\ \Omega + 4\ \Omega = 8\ \Omega\$
Battery current (Ohm’s law):
\$I{\text{total}} = \frac{V}{R{\text{total}}}= \frac{12\ \text{V}}{8\ \Omega}=1.5\ \text{A}\$
Voltage across the parallel section (series rule):
\$V{23}= I{\text{total}} \times R_{23}=1.5\ \text{A}\times4\ \Omega=6\ \text{V}\$
Using the junction rule:
\$I2 = \frac{V{23}}{R_2}= \frac{6\ \text{V}}{6\ \Omega}=1.0\ \text{A}\$
\$I3 = \frac{V{23}}{R_3}= \frac{6\ \text{V}}{12\ \Omega}=0.5\ \text{A}\$
Check: \$I2+I3 = 1.0\ \text{A}+0.5\ \text{A}=1.5\ \text{A}=I_{\text{total}}\$ (junction rule satisfied).
Voltage across each resistor:
Note that the parallel branches each have the same voltage (6 V), confirming the parallel voltage rule.
Parallel equivalent: \$\frac{1}{R{\text{eq}}}= \frac{1}{3}+ \frac{1}{6}= \frac{2}{6}+ \frac{1}{6}= \frac{3}{6}\$, so \$R{\text{eq}}=2\ \Omega\$.
Total current: \$I_{\text{total}} = \frac{9\ \text{V}}{2\ \Omega}=4.5\ \text{A}\$.
\$I1 = \frac{9\ \text{V}}{3\ \Omega}=3.0\ \text{A}\$, \$I2 = \frac{9\ \text{V}}{6\ \Omega}=1.5\ \text{A}\$ (junction rule check: \$3.0+1.5=4.5\ \text{A}\$).
Total resistance \$R_{\text{total}} = 2+4+8 = 14\ \Omega\$.
Current \$I = \frac{12\ \text{V}}{14\ \Omega}=0.857\ \text{A}\$.
\$V1 = I R1 =0.857\times2=1.71\ \text{V}\$,
\$V2 = I R2 =0.857\times4=3.43\ \text{V}\$,
\$V3 = I R3 =0.857\times8=6.86\ \text{V}\$ (sum ≈12 V).
Parallel part: \$\frac{1}{R{p}}= \frac{1}{10}+ \frac{1}{15}= \frac{3}{30}+ \frac{2}{30}= \frac{5}{30}\$, so \$R{p}=6\ \Omega\$.
Total resistance \$R_{\text{total}} =5+6=11\ \Omega\$.
Total current \$I = \frac{24\ \text{V}}{11\ \Omega}=2.18\ \text{A}\$ (flows through the 5 Ω resistor).
Voltage across the parallel network \$V_{p}= I \times 5\ \Omega = 10.9\ \text{V}\$.
Current through 10 Ω resistor \$I{10}= \frac{V{p}}{10}=1.09\ \text{A}\$, voltage across it \$=10.9\ \text{V}\$ (same as \$V_{p}\$, confirming the parallel voltage rule).