Published by Patrick Mutisya · 14 days ago
Derive, using Kirchhoff’s laws, a formula for the combined resistance of two or more resistors connected in series.
\$\sum I{\text{in}} = \sum I{\text{out}}\$
\$\sum V = 0\$
Consider a simple series circuit powered by a single ideal battery of emf \$V\$ and internal resistance negligible. The circuit contains \$n\$ resistors \$R1, R2, \dots , R_n\$ as shown in the suggested diagram.
\$V - V1 - V2 - \dots - V_n = 0.\$
\$V - I R1 - I R2 - \dots - I R_n = 0.\$
\$V = I\,(R1 + R2 + \dots + R_n).\$
\$R{\text{eq}} = R1 + R2 + \dots + Rn = \sum{k=1}^{n} Rk.\$
The combined resistance of resistors connected in series is simply the arithmetic sum of the individual resistances:
\$\boxed{R{\text{eq}} = \sum{k=1}^{n} R_k}\$
| Law | Mathematical Form | Physical Meaning |
|---|---|---|
| Kirchhoff’s Current Law (KCL) | \$\displaystyle\sum I{\text{in}} = \sum I{\text{out}}\$ | Charge is conserved at a junction; no accumulation of charge. |
| Kirchhoff’s \cdot oltage Law (K \cdot L) | \$\displaystyle\sum V = 0\$ | The net change in electric potential around a closed loop is zero. |
| Series Resistance | \$\displaystyle R{\text{eq}} = \sum{k=1}^{n} R_k\$ | Resistors share the same current; total voltage drop is the sum of individual drops. |
Using the same method, derive the formula for the equivalent resistance of two resistors \$R1\$ and \$R2\$ connected in parallel.