Mastering Series and Parallel Combinations of Resistors for JEE Physics
Understanding how resistors combine in series and parallel circuits is fundamental for solving complex electrical problems in competitive exams like JEE. This module will break down these concepts, providing clear explanations, practical examples, and opportunities for active recall.
Series Combinations
In a series combination, resistors are connected end-to-end, forming a single path for current to flow. The current through each resistor is the same, while the total voltage across the combination is the sum of the voltages across individual resistors.
In series, current is constant, voltage adds up.
When resistors are in series, the total resistance is simply the sum of individual resistances. This means adding more resistors in series will always increase the total resistance.
For resistors R₁, R₂, R₃, ..., R<0xE2><0x82><0x99> connected in series, the equivalent resistance (R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0>) is given by: R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0> = R₁ + R₂ + R₃ + ... + R<0xE2><0x82><0x99> The current (I) flowing through the circuit is the same for all resistors. The total voltage (V) across the series combination is the sum of the voltage drops across each resistor: V = V₁ + V₂ + V₃ + ... + V<0xE2><0x82><0x99>. Using Ohm's Law (V=IR), we can see that V = IR₁ + IR₂ + IR₃ + ... + IR<0xE2><0x82><0x99> = I(R₁ + R₂ + R₃ + ... + R<0xE2><0x82><0x99>). Since V = IR<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0>, it follows that R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0> = R₁ + R₂ + R₃ + ... + R<0xE2><0x82><0x99>.
15Ω (5Ω + 10Ω)
Parallel Combinations
In a parallel combination, resistors are connected across the same two points, providing multiple paths for current to flow. The voltage across each resistor is the same, while the total current is the sum of the currents through individual resistors.
In parallel circuits, the voltage across each component is identical. The total current entering the parallel combination splits among the branches, and the sum of these branch currents equals the total current. The reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. This means adding more resistors in parallel will always decrease the total resistance, as it provides more pathways for current.
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In parallel, voltage is constant, current adds up.
For resistors R₁, R₂, R₃, ..., R<0xE2><0x82><0x99> connected in parallel, the reciprocal of the equivalent resistance (R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0>) is the sum of the reciprocals of the individual resistances: 1/R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0> = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/R<0xE2><0x82><0x99>.
For two resistors R₁ and R₂ in parallel, the equivalent resistance can be simplified to: R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0> = (R₁ * R₂) / (R₁ + R₂). The voltage (V) across each resistor is the same. The total current (I) is the sum of the currents through each resistor: I = I₁ + I₂ + I₃ + ... + I<0xE2><0x82><0x99>. Using Ohm's Law (I=V/R), we get I = V/R₁ + V/R₂ + V/R₃ + ... + V/R<0xE2><0x82><0x99> = V(1/R₁ + 1/R₂ + 1/R₃ + ... + 1/R<0xE2><0x82><0x99>). Since I = V/R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0>, it follows that 1/R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0> = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/R<0xE2><0x82><0x99>.
2Ω ( (6*3) / (6+3) = 18 / 9 = 2 )
Mixed Combinations and Problem Solving
Many JEE problems involve circuits with resistors arranged in both series and parallel configurations. The key to solving these is to simplify the circuit step-by-step, identifying series and parallel groups and replacing them with their equivalent resistances until a single equivalent resistance is found.
| Feature | Series Combination | Parallel Combination |
|---|---|---|
| Current | Same through all resistors | Splits among resistors |
| Voltage | Adds up across resistors | Same across all resistors |
| Equivalent Resistance | R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0> = R₁ + R₂ + ... | 1/R<0xE2><0x82><0x91><0xE1><0xB5><0xA0><0xE1><0xB5><0xA0> = 1/R₁ + 1/R₂ + ... |
| Effect of adding more resistors | Increases total resistance | Decreases total resistance |
Analogy: Think of series resistors like a single-lane road where all cars must pass through each toll booth. Parallel resistors are like a multi-lane highway where cars can choose different lanes, reducing overall congestion.
Key Takeaways for JEE
Always identify the simplest series or parallel combinations first. Pay close attention to the direction of current and voltage drops. Practice with a variety of circuit diagrams to build confidence and speed.
Learning Resources
Provides a clear, foundational explanation of resistance and how resistors behave in series and parallel circuits.
Detailed explanation of series and parallel resistor combinations with formulas and conceptual understanding.
A video tutorial specifically tailored for JEE preparation, covering problem-solving techniques for resistor combinations.
Explains the concept of equivalent resistance and provides interactive examples for series and parallel circuits.
Offers practice problems and solutions related to series and parallel resistor combinations, useful for JEE aspirants.
A comprehensive guide to understanding the behavior of components in series and parallel circuits.
Another excellent video resource focusing on JEE Mains preparation for current electricity, with a dedicated section on resistor combinations.
Provides a broad overview and mathematical derivations for series and parallel circuit configurations.
A practical guide with clear steps and examples on calculating equivalent resistance in various circuit configurations.
A collection of practice problems with solutions for current electricity, including many on series and parallel resistor combinations.