⚡ KCL (Kirchhoff’s Current Law) says that the total current entering a junction equals the total current leaving it.
⚡ KVL (Kirchhoff’s Voltage Law) says that the sum of all voltage drops around any closed loop is zero.
Think of a river splitting into two streams (currents) that later join again – the amount of water that leaves the junction equals the amount that enters.
Consider a simple circuit with a voltage source \(V\) connected to two resistors \(R1\) and \(R2\) in parallel (see the diagram below).
By KVL, the voltage across each resistor is the same as the source voltage:
\$V{R1}=V{R2}=V\$
By Ohm’s law, the currents through the resistors are
\$I1=\frac{V}{R1},\qquad I2=\frac{V}{R2}\$
Apply KCL at the node where the two currents join:
\$I{\text{total}} = I1 + I2 = \frac{V}{R1} + \frac{V}{R_2}\$
Define the total resistance \(R{\text{eq}}\) of the parallel network by \(I{\text{total}} = \frac{V}{R_{\text{eq}}}\).
Equate the two expressions for \(I_{\text{total}}\):
\$\frac{V}{R{\text{eq}}} = \frac{V}{R1} + \frac{V}{R_2}\$
Cancel the common voltage \(V\) (assuming \(V \neq 0\)) and invert to get the familiar formula:
\$\boxed{\frac{1}{R{\text{eq}}} = \frac{1}{R1} + \frac{1}{R_2}}\$
For more than two resistors, the same reasoning gives
\$\boxed{\frac{1}{R{\text{eq}}} = \sum{i=1}^{n}\frac{1}{R_i}}\$
Let’s calculate the equivalent resistance for \(R1=10\,\Omega\) and \(R2=20\,\Omega\).
| Resistor | Resistance (Ω) |
|---|---|
| \(R_1\) | 10 |
| \(R_2\) | 20 |
| Result | \$R_{\text{eq}} = \frac{1}{\frac{1}{10}+\frac{1}{20}} = \frac{1}{0.1+0.05} = \frac{1}{0.15} \approx 6.67\,\Omega\$ |
When resistors are connected side‑by‑side (parallel), the total resistance is always less than the smallest individual resistance. This is because the current has multiple paths to flow, just like water flowing through several pipes at once.