The Equivalent Resistance Formula gives the single resistance value that can replace a combination of resistors in a circuit without changing the current or voltage behaviour. This formula is a core concept in NCERT Class 10 Physics (Chapter 12 — Electricity) and is revisited in Class 12. It is equally important for JEE Main, JEE Advanced, and NEET, where circuit analysis questions appear every year. This article covers the formula for series and parallel combinations, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Equivalent Resistance Formulas at a Glance
Quick reference for the most important equivalent resistance formulas.
- Series combination: \( R_{eq} = R_1 + R_2 + R_3 + \cdots + R_n \)
- Parallel combination: \( \dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \cdots + \dfrac{1}{R_n} \)
- Two resistors in parallel: \( R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2} \)
- n equal resistors in series: \( R_{eq} = nR \)
- n equal resistors in parallel: \( R_{eq} = \dfrac{R}{n} \)
- Ohm's Law: \( V = IR \)
- Power dissipated: \( P = \dfrac{V^2}{R} = I^2 R \)
What is the Equivalent Resistance Formula?
The Equivalent Resistance Formula defines the total resistance of a network of resistors as a single equivalent value. When resistors are connected in a circuit, they can be arranged in two fundamental ways — series and parallel. In a series combination, the same current flows through every resistor. The equivalent resistance is simply the sum of all individual resistances. In a parallel combination, the same voltage appears across every resistor. The reciprocal of the equivalent resistance equals the sum of the reciprocals of individual resistances.
This concept is introduced in NCERT Class 10 Science, Chapter 12 (Electricity). It is further explored in NCERT Class 12 Physics, Chapter 3 (Current Electricity). Understanding this formula helps students analyse complex circuits, design electrical systems, and solve numerical problems in board exams and competitive tests. The equivalent resistance concept underpins Kirchhoff's Laws, Wheatstone Bridge, and potentiometer circuits.
Equivalent Resistance Formula — Expression and Variables
Series Combination
When resistors \( R_1, R_2, R_3, \ldots, R_n \) are connected end-to-end (in series):
\[ R_{eq} = R_1 + R_2 + R_3 + \cdots + R_n \]
Parallel Combination
When resistors \( R_1, R_2, R_3, \ldots, R_n \) are connected across the same two nodes (in parallel):
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots + \frac{1}{R_n} \]
For exactly two resistors in parallel, this simplifies to:
\[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( R_{eq} \) | Equivalent (total) resistance | Ohm (Ω) |
| \( R_1, R_2, \ldots, R_n \) | Individual resistances | Ohm (Ω) |
| \( I \) | Current through the circuit | Ampere (A) |
| \( V \) | Voltage (potential difference) | Volt (V) |
| \( n \) | Number of resistors | Dimensionless |
Derivation — Series Combination
Consider three resistors \( R_1 \), \( R_2 \), and \( R_3 \) connected in series across a battery of voltage \( V \).
Step 1: The same current \( I \) flows through each resistor. By Ohm's Law, the voltage drops are \( V_1 = IR_1 \), \( V_2 = IR_2 \), and \( V_3 = IR_3 \).
Step 2: The total voltage equals the sum of individual drops: \( V = V_1 + V_2 + V_3 \).
Step 3: Substituting: \( IR_{eq} = IR_1 + IR_2 + IR_3 \). Dividing throughout by \( I \) gives \( R_{eq} = R_1 + R_2 + R_3 \).
Derivation — Parallel Combination
Consider three resistors \( R_1 \), \( R_2 \), and \( R_3 \) connected in parallel across voltage \( V \).
Step 1: The same voltage \( V \) appears across each resistor. By Ohm's Law, the branch currents are \( I_1 = V/R_1 \), \( I_2 = V/R_2 \), and \( I_3 = V/R_3 \).
Step 2: The total current is \( I = I_1 + I_2 + I_3 \).
Step 3: Since \( I = V/R_{eq} \), substituting gives \( V/R_{eq} = V/R_1 + V/R_2 + V/R_3 \). Dividing by \( V \) yields \( 1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3 \).
Complete Electricity Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Equivalent Resistance (Series) | \( R_{eq} = R_1 + R_2 + \cdots + R_n \) | R = resistance of each resistor | Ω | Class 10, Ch 12 / Class 12, Ch 3 |
| Equivalent Resistance (Parallel) | \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} \) | R = resistance of each resistor | Ω | Class 10, Ch 12 / Class 12, Ch 3 |
| Two Resistors in Parallel | \( R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \) | R₁, R₂ = individual resistances | Ω | Class 10, Ch 12 |
| Ohm's Law | \( V = IR \) | V = voltage, I = current, R = resistance | V, A, Ω | Class 10, Ch 12 |
| Electrical Resistance | \( R = \frac{\rho L}{A} \) | ρ = resistivity, L = length, A = area | Ω | Class 12, Ch 3 |
| Electric Power | \( P = VI = I^2 R = \frac{V^2}{R} \) | P = power, V = voltage, I = current | W | Class 10, Ch 12 |
| Electrical Energy | \( E = Pt = VIt \) | E = energy, P = power, t = time | J (or kWh) | Class 10, Ch 12 |
| Conductance | \( G = \frac{1}{R} \) | G = conductance, R = resistance | Siemens (S) | Class 12, Ch 3 |
| Wheatstone Bridge Condition | \( \frac{P}{Q} = \frac{R}{S} \) | P, Q, R, S = resistances in bridge arms | Ω | Class 12, Ch 3 |
| n Equal Resistors in Series | \( R_{eq} = nR \) | n = number of resistors, R = each resistance | Ω | Class 10, Ch 12 |
| n Equal Resistors in Parallel | \( R_{eq} = \frac{R}{n} \) | n = number of resistors, R = each resistance | Ω | Class 10, Ch 12 |
Equivalent Resistance Formula — Solved Examples
Example 1 (Class 9-10 Level) — Series Combination
Problem: Three resistors of 4 Ω, 6 Ω, and 10 Ω are connected in series across a 20 V battery. Find the equivalent resistance and the current flowing through the circuit.
Given: \( R_1 = 4\,\Omega \), \( R_2 = 6\,\Omega \), \( R_3 = 10\,\Omega \), \( V = 20\,\text{V} \)
Step 1: Apply the series formula: \( R_{eq} = R_1 + R_2 + R_3 \)
Step 2: Substitute values: \( R_{eq} = 4 + 6 + 10 = 20\,\Omega \)
Step 3: Use Ohm's Law to find current: \( I = \dfrac{V}{R_{eq}} = \dfrac{20}{20} = 1\,\text{A} \)
Answer
Equivalent resistance = 20 Ω; Current = 1 A
Example 2 (Class 11-12 Level) — Mixed Series-Parallel Combination
Problem: Two resistors of 6 Ω and 3 Ω are connected in parallel. This parallel combination is then connected in series with a 4 Ω resistor. The entire network is connected to a 16 V supply. Find (a) the equivalent resistance of the network, (b) the total current drawn from the supply, and (c) the current through each parallel branch.
Given: \( R_A = 6\,\Omega \), \( R_B = 3\,\Omega \) (in parallel); \( R_C = 4\,\Omega \) (in series); \( V = 16\,\text{V} \)
Step 1: Find the equivalent resistance of the parallel combination:
\( R_{AB} = \dfrac{R_A \times R_B}{R_A + R_B} = \dfrac{6 \times 3}{6 + 3} = \dfrac{18}{9} = 2\,\Omega \)
Step 2: Add the series resistor:
\( R_{eq} = R_{AB} + R_C = 2 + 4 = 6\,\Omega \)
Step 3: Find the total current using Ohm's Law:
\( I = \dfrac{V}{R_{eq}} = \dfrac{16}{6} \approx 2.67\,\text{A} \)
Step 4: Find voltage across the parallel combination:
\( V_{AB} = I \times R_{AB} = 2.67 \times 2 \approx 5.33\,\text{V} \)
Step 5: Find current through each branch:
\( I_A = \dfrac{V_{AB}}{R_A} = \dfrac{5.33}{6} \approx 0.89\,\text{A} \)
\( I_B = \dfrac{V_{AB}}{R_B} = \dfrac{5.33}{3} \approx 1.78\,\text{A} \)
Answer
(a) Equivalent resistance = 6 Ω; (b) Total current ≈ 2.67 A; (c) Current through 6 Ω branch ≈ 0.89 A, current through 3 Ω branch ≈ 1.78 A
Example 3 (JEE/NEET Level) — Infinite Ladder Network
Problem: An infinite ladder network consists of resistors \( R_1 = 2\,\Omega \) (series arm) and \( R_2 = 2\,\Omega \) (shunt arm) repeating indefinitely. Find the equivalent resistance \( R_{eq} \) of the infinite ladder between the input terminals.
Given: \( R_1 = 2\,\Omega \), \( R_2 = 2\,\Omega \), infinite repetitions.
Step 1: For an infinite network, adding one more section does not change the equivalent resistance. Let the equivalent resistance be \( R_{eq} \).
Step 2: The input resistance equals \( R_1 \) in series with the parallel combination of \( R_2 \) and \( R_{eq} \):
\( R_{eq} = R_1 + \dfrac{R_2 \cdot R_{eq}}{R_2 + R_{eq}} \)
Step 3: Substitute \( R_1 = 2 \) and \( R_2 = 2 \):
\( R_{eq} = 2 + \dfrac{2 R_{eq}}{2 + R_{eq}} \)
Step 4: Multiply both sides by \( (2 + R_{eq}) \):
\( R_{eq}(2 + R_{eq}) = 2(2 + R_{eq}) + 2R_{eq} \)
\( 2R_{eq} + R_{eq}^2 = 4 + 2R_{eq} + 2R_{eq} \)
\( R_{eq}^2 = 4 + 2R_{eq} \)
\( R_{eq}^2 – 2R_{eq} – 4 = 0 \)
Step 5: Solve the quadratic equation:
\( R_{eq} = \dfrac{2 + \sqrt{4 + 16}}{2} = \dfrac{2 + \sqrt{20}}{2} = 1 + \sqrt{5} \approx 3.24\,\Omega \)
(Taking the positive root since resistance cannot be negative.)
Answer
Equivalent resistance of the infinite ladder = \( (1 + \sqrt{5})\,\Omega \approx 3.24\,\Omega \)
CBSE Exam Tips 2025-26
- Draw the circuit first. Always sketch the resistor network before solving. This prevents errors in identifying series and parallel branches. We recommend this as your first step in every circuit problem.
- Simplify step by step. Reduce the most deeply nested combination first. Work outward towards the terminals. Do not try to solve the entire circuit in one step.
- Remember the key rule: In series, \( R_{eq} \) is always greater than the largest individual resistance. In parallel, \( R_{eq} \) is always smaller than the smallest individual resistance. Use this to check your answer.
- Show your formula. In CBSE 2025-26 board exams, writing the formula before substituting values earns you step marks even if the final answer is wrong.
- Use the product-over-sum shortcut for two resistors in parallel: \( R_{eq} = R_1 R_2 / (R_1 + R_2) \). This saves time in 1-mark and 2-mark questions.
- Watch for trick questions where a wire (zero resistance) is placed in parallel with a resistor. The equivalent resistance of that combination is zero.
Common Mistakes to Avoid
- Mistake 1 — Adding reciprocals in series: Some students write \( 1/R_{eq} = 1/R_1 + 1/R_2 \) for a series circuit. This is the parallel formula. For series, simply add the resistances directly.
- Mistake 2 — Forgetting to take the reciprocal: After calculating \( 1/R_{eq} \) for a parallel combination, students often forget to invert the result. Always write \( R_{eq} = 1 / (1/R_1 + 1/R_2) \) and compute the final value.
- Mistake 3 — Misidentifying series vs. parallel: Two resistors are in parallel only if both their terminals are connected to the same two nodes. Trace the circuit carefully. Redraw if necessary.
- Mistake 4 — Ignoring internal resistance: In problems involving a battery with internal resistance \( r \), the internal resistance is in series with the external circuit. Always include it in the series sum.
- Mistake 5 — Wrong units: Always express all resistances in Ohms (Ω) before substituting. Mixing kΩ and Ω leads to incorrect answers in numerical problems.
JEE/NEET Application of the Equivalent Resistance Formula
In our experience, JEE aspirants encounter the Equivalent Resistance Formula in at least 2-3 questions per paper. NEET also tests this concept in the context of biological systems (nerve impulse resistance models) and standard circuit problems. Here are the key application patterns you must master.
Pattern 1 — Wheatstone Bridge
The Wheatstone Bridge is a balanced resistor network. When \( P/Q = R/S \), no current flows through the galvanometer. The bridge then reduces to two series pairs connected in parallel. You apply the Equivalent Resistance Formula twice — once for each series pair, then once for the resulting parallel combination. JEE often asks you to find the equivalent resistance of an unbalanced bridge, requiring you to apply Kirchhoff's Laws along with the formula.
Pattern 2 — Symmetry-Based Simplification
JEE Advanced frequently presents cube or hexagonal resistor networks. The key insight is that points at equal potential can be joined (short-circuited) or separated (open-circuited) without affecting the circuit. After applying symmetry, the complex network reduces to a simple series-parallel combination. Our experts suggest practising the cube network (12 equal resistors) as it appears almost every alternate year in JEE Advanced.
Pattern 3 — Infinite Networks
As shown in Example 3, infinite ladder networks are solved by assuming the equivalent resistance equals itself after one additional section. This leads to a quadratic equation. The positive root gives the answer. JEE Main has tested this concept multiple times. The key step is setting up the self-similar equation correctly.
Pattern 4 — Variable Resistance and Graphs
NEET and JEE Main ask questions where one resistor in a parallel combination changes (e.g., a thermistor or LDR). You must express \( R_{eq} \) as a function of the variable resistance and analyse the resulting graph. Understanding how \( R_{eq} \) varies with individual resistances is essential for these problems.
FAQs on Equivalent Resistance Formula
Explore More Physics Formulas
Now that you have mastered the Equivalent Resistance Formula, strengthen your Physics preparation with these related resources on ncertbooks.net:
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For the official NCERT curriculum, refer to the NCERT official website to access the latest Class 10 and Class 12 Physics textbooks.