derive, using Kirchhoff’s laws, a formula for the combined resistance of two or more resistors in parallel

Kirchhoff’s Laws – Derivation of Equivalent Resistance

Objective

Using Kirchhoff’s Current Law (KCL), Kirchhoff’s Voltage Law (KVL) and Ohm’s law, derive the formulas for the equivalent resistance of:

  • Resistors in series

  • Resistors in parallel (two or more branches)

  • Resistors forming a bridge (Wheatstone‑type circuit)

Apply the results to typical A‑Level problems such as potential dividers, loaded dividers and bridge balance.

Key Concepts

  • KCL: The algebraic sum of currents at a junction is zero.


    Taking currents entering the node as positive,

    \$\sum I = 0 \;\Longrightarrow\; I{\text{in}} = I{\text{out}}.\$

  • KVL: The algebraic sum of potential differences around any closed loop is zero.

    \$\sum V = 0.\$

  • Ohm’s law: For a resistor, \$V = IR\$ (or \$I = V/R\$).

1. Series Resistors – Derivation

  1. Consider a single loop containing a source of emf \$V\$ and \$n\$ resistors \$R1,R2,\dots,R_n\$ in series.

  2. Apply KVL around the loop:

    \$V - IR1 - IR2 -\dots -IR_n = 0,\$

    where the same current \$I\$ flows through every element (KCL guarantees a single current in a series chain).

  3. Re‑arrange:

    \$V = I\,(R1+R2+\dots+R_n).\$

  4. Define the equivalent resistance \$R{\text{eq}}\$ by \$V = IR{\text{eq}}\$. Hence

    \$\boxed{R{\text{eq}} = R1+R2+\dots+Rn}.\$

2. Parallel Resistors – Derivation

2.1 Two Resistors in Parallel

  1. Diagram: a voltage source \$V\$ feeds node A; from A two branches contain \$R1\$ and \$R2\$ and re‑join at node B.

  2. KCL at node A (currents leaving the node taken as positive):

    \$I{\!s}=I1+I_2.\$

  3. Ohm’s law for each branch (the voltage across each resistor is the same, \$V\$):

    \$I1=\frac{V}{R1},\qquad I2=\frac{V}{R2}.\$

  4. Substitute into the KCL equation:

    \$I{\!s}= \frac{V}{R1}+\frac{V}{R2}=V\!\left(\frac1{R1}+\frac1{R_2}\right).\$

  5. Define \$R{\text{eq}}\$ of the parallel network by \$I{\!s}=V/R_{\text{eq}}\$ and equate:

    \$\frac{1}{R{\text{eq}}}= \frac1{R1}+\frac1{R_2}.\$

    \$\boxed{R{\text{eq}}=\frac{1}{\displaystyle\frac1{R1}+\frac1{R_2}}}.\$

2.2 Generalisation to \$n\$ Parallel Resistors

  1. For \$n\$ branches each carries \$Ik = V/Rk\$ (same \$V\$ across all).

  2. KCL at the common node gives

    \$I{\!s}= \sum{k=1}^{n} Ik = V\sum{k=1}^{n}\frac1{R_k}.\$

  3. With \$I{\!s}=V/R{\text{eq}}\$, we obtain

    \$\boxed{\frac1{R{\text{eq}}}= \sum{k=1}^{n}\frac1{R_k}}\$

    or equivalently

    \$\boxed{R{\text{eq}}=\left(\displaystyle\sum{k=1}^{n}\frac1{R_k}\right)^{-1}}.\$

2.3 Bridge (Wheatstone) Circuit

Consider the classic bridge shown below, with resistors \$R1,R2,R3,R4\$ forming the four arms and a galvanometer (or load) of resistance \$R_5\$ across the bridge points C and D.

  1. Apply KVL to the two loops on either side of the bridge:

    • Loop ABCD: \$V{AB}=I1R1+I5R5+I3R_3\$

    • Loop BCDA: \$V{AB}=I2R2+I5R5+I4R_4\$

  2. KCL at nodes C and D gives

    \$I1=I3+I5,\qquad I2=I4+I5.\$

  3. Eliminate the bridge current \$I_5\$ by subtracting the two KVL equations:

    \$I1R1+I3R3 = I2R2+I4R4.\$

    Using the KCL relations this reduces to the *balance condition*

    \$\boxed{\frac{R1}{R2}= \frac{R3}{R4}}\$

    (when \$I_5=0\$, i.e. the bridge is balanced).

  4. If the bridge is unbalanced, replace the two parallel arms \$(R1\parallel R2)\$ and \$(R3\parallel R4)\$ by their equivalents, then combine with \$R_5\$ using the parallel formulas derived above. The total equivalent resistance seen by the source is

    \$R{\text{eq}} = \bigl(R1\parallel R2\bigr)+\bigl(R3\parallel R4\bigr)+R5,\$

    where \$Ra\parallel Rb = \dfrac{RaRb}{Ra+Rb}.\$

3. Practical Notes – Real‑World Effects (10.1 Practical Circuits)

Remember when moving from ideal theory to the laboratory:

  • Wire resistance: Even short leads have a small resistance; in precision work they are often included as series resistors.

  • Internal resistance of sources: A real battery or power supply can be modelled as an ideal emf \$V\$ in series with an internal resistance \$r\$. KVL then becomes \$V - Ir - \sum V_{\text{drops}} = 0\$.

  • Polarity and connection errors: Reversing the polarity of a source changes the sign of the voltage term in KVL.

  • Instrument loading: A voltmeter has a finite internal resistance \$R_{\text{vm}}\$; when connected across a component it forms a parallel combination, altering the measured voltage.

  • Measurement uncertainty: Include the tolerance of resistors, the accuracy class of ammeters/voltmeters, and contact resistance in error analysis.

4. Potential Dividers – Application (10.3)

4.1 Unloaded Divider

Two series resistors \$R{\text{top}}\$ and \$R{\text{bottom}}\$ are connected across a source \$V_s\$. The output voltage at the junction is

\$V{\text{out}} = Vs\;\frac{R{\text{bottom}}}{R{\text{top}}+R_{\text{bottom}}}.\$

4.2 Loaded Divider (Resistor \$R_L\$ across the lower leg)

  1. Replace the lower leg by the parallel combination \$R{\text{b}} = R{\text{bottom}}\parallel R_L\$:

    \$\frac1{R{\text{b}}}= \frac1{R{\text{bottom}}}+ \frac1{R_L}.\$

  2. The total series resistance seen by the source is \$R{\text{tot}} = R{\text{top}}+R_{\text{b}}\$.

  3. Current from the source:

    \$I = \frac{Vs}{R{\text{tot}}}.\$

  4. Voltage across the parallel pair (and therefore across \$R_L\$) is

    \$V{\text{out}} = I\,R{\text{b}} = Vs\;\frac{R{\text{b}}}{R{\text{top}}+R{\text{b}}}.\$

    Substituting \$R_{\text{b}}\$ gives the familiar loaded‑divider formula

    \$\boxed{V{\text{out}} = Vs\;\frac{R{\text{bottom}}\,RL}{R{\text{top}}(R{\text{bottom}}+RL)+R{\text{bottom}}\,R_L}}.\$

5. Summary Table

ConfigurationEquivalent‑Resistance FormulaKey Steps Using Kirchhoff’s Laws
Series (any number)\$R{\text{eq}} = R1+R2+\dots+Rn\$KVL around the loop → factor out common current \$I\$ → solve for \$R_{\text{eq}}\$.
Two resistors in parallel\$\displaystyle R{\text{eq}} = \frac{1}{\frac1{R1}+\frac1{R_2}}\$KCL at the junction → Ohm’s law for each branch → combine.
\$n\$ resistors in parallel\$\displaystyle \frac1{R{\text{eq}}}= \sum{k=1}^{n}\frac1{R_k}\$Generalise KCL to \$n\$ branches → apply Ohm’s law → sum and invert.
Wheatstone bridge (balanced)\$\displaystyle \frac{R1}{R2}= \frac{R3}{R4}\$KVL on the two loops + KCL at the bridge nodes → eliminate bridge current.
Unbalanced bridge (general case)\$R{\text{eq}} = (R1\parallel R2)+(R3\parallel R4)+R5\$Use parallel formula for each arm, then add series resistances.

6. Connections to Other Syllabus Topics

  • 10.2 Kirchhoff’s laws: Derivations above cover series, parallel and bridge circuits; the balance condition is a classic exam question.

  • 10.3 Potential dividers: Both unloaded and loaded cases are solved using the parallel‑resistance result.

  • 10.1 Practical circuits: The “Practical notes” box reminds students of internal resistance, wire resistance and instrument loading.

  • 20.5 Electromagnetic induction: When analysing induced‑emf loops, the same KVL + equivalent‑resistance approach is used, e.g. in a loop containing a coil and a parallel resistor.

7. Suggested Diagram

Figure 1: (a) Two resistors \$R1,R2\$ in parallel fed by a source \$V\$. (b) Wheatstone bridge with resistors \$R1\$–\$R4\$ and bridge resistor \$R_5\$.