Table of Contents
In a parallel connection, all component terminals share the same two node potentials. By the node rule, the total current $I$ splits into branch currents $I_1,\dots,I_n$. Dividing by the applied voltage $U$ gives
\[
\frac{I}{U}=\frac{I_1}{U}+\frac{I_2}{U}+\dots+\frac{I_n}{U}.
\]
Resistors in parallel
Using Ohm’s law in each branch, each term is the reciprocal of a branch resistance. Hence, for $n$ resistors,
\[
\boxed{\frac{1}{R_\mathrm{tot}}=\sum_{i=1}^n \frac{1}{R_i}}.
\]
Intuitively, adding parallel paths increases the effective cross-section, so the total resistance decreases.
Capacitors in parallel
Total charge adds:
\[
Q=Q_1+Q_2+\dots+Q_n.
\]
Divide by $U$ and use $C=Q/U$ to get
\[
\boxed{C_\mathrm{tot}=\sum_{i=1}^n C_i}.
\]
This is analogous to enlarging plate area, so total capacitance increases.
Inductors in parallel
Differentiate the current sum and divide by $U$; with $U = L\,\dot I$ (per branch) you obtain
\[
\boxed{\frac{1}{L_\mathrm{tot}}=\sum_{i=1}^n \frac{1}{L_i}}.
\]
Example
In three separate parallel circuits, find:
- a) Total resistance of $100\,\Omega$ and $20\,\Omega$:
\[
\frac{1}{R_\mathrm{tot}}=\frac{1}{100}+\frac{1}{20}=0.01+0.05=0.06
\Rightarrow R_\mathrm{tot}=16.7\,\Omega.
\] - b) Total capacitance of $50\,\mu\text{F}$, $10\,\mu\text{F}$, and $250\,\text{nF}$:
\[
C_\mathrm{tot}=50\,\mu\text{F}+10\,\mu\text{F}+0.25\,\mu\text{F}=60.25\,\mu\text{F}.
\] - c) Total inductance of $30\,\text{mH}$, $120\,\text{mH}$, and $240\,\text{mH}$:
\[
\frac{1}{L_\mathrm{tot}}=\frac{1}{30}+\frac{1}{120}+\frac{1}{240}=0.045833\;\text{mH}^{-1}
\Rightarrow L_\mathrm{tot}\approx 21.8\,\text{mH}.
\]