Resistors in series & parallel

Resistors in series & parallel

Here is a comparison of Series and Parallel circuits to get us started. This table summarises the most important differences you'll need for your exams.

Understanding these fundamental differences is the key to solving any problem involving resistor combinations. We will now break down each type in detail, with plenty of solved numerical examples to build your confidence.

Resistors in Series: The Single Path

When components are connected end-to-end, creating a single, unbroken loop for the current to flow, they are said to be in series. Think of it like cars on a one-lane road; they all have to follow the same path and travel at the same rate (current).

The key characteristics of a series circuit are:

1. Same Current: The electric current (I) is identical through every resistor in the series. I_total = I₁ = I₂ = I₃ = ...
2. Voltage Division: The total voltage supplied by the source (V_total) is divided among the resistors. The sum of the voltage drops across each resistor equals the total voltage. V_total = V₁ + V₂ + V₃ + ...

{{VISUAL: diagram: A simple circuit diagram showing a battery connected to three resistors (R₁, R₂, R₃) in a single loop, illustrating a series connection. Current (I) is shown flowing from the positive to the negative terminal through all resistors.}}

Deriving the Formula for Equivalent Resistance

The equivalent resistance (R_eq or R_series) is the single resistance value that could replace the entire series combination without changing the total current or voltage in the circuit.

Using Ohm's Law (V = I × R) and our two rules for series circuits:

1. We know V_total = V₁ + V₂ + V₃ + ...
2. Substitute V with I × R: I_total × R_eq = (I₁ × R₁) + (I₂ × R₂) + (I₃ × R₃) + ...
3. Since the current is the same everywhere (I_total = I₁ = I₂ = ...), we can cancel I from every term.

This leaves us with the fundamental formula for resistors in series.

{FORMULA: expr=Rₑ = R₁ + R₂ + R₃ + ... + Rₙ | symbols=Rₑ: Equivalent Resistance (Ω), R₁, R₂, ...: Individual Resistances (Ω)}

This formula tells us something very important: adding a resistor in series always increases the total resistance of the circuit. This makes sense, as you are adding more obstacles to the single path of the current.

{CALLOUT: type=tip | text=Memory Hook for Series: Think Series, Single path, Sum. The equivalent resistance is the simple sum of the individual resistances.}

Solved Example 1: Basic Series Circuit (Easy)

Given: Three resistors R₁ = 5 Ω, R₂ = 10 Ω, and R₃ = 15 Ω are connected in series to a 12 V battery.

To Find: (a) The total equivalent resistance (Rₑ). (b) The total current (I) flowing from the battery.

Solution:

1. Find the equivalent resistance (Rₑ). Since the resistors are in series, we use the sum formula.

   Rₑ = R₁ + R₂ + R₃
   

   Rₑ = 5 Ω + 10 Ω + 15 Ω = 30 Ω
   

2. Find the total current (I). Now, we can treat the entire circuit as a single 30 Ω resistor connected to a 12 V battery. We use Ohm's Law, V = I × R.

   I = V / Rₑ
   

   I = 12 V / 30 Ω = 0.4 A
   

Final Answer: (a) The equivalent resistance is 30 Ω. (b) The total current is 0.4 A.

Resistors in Parallel: Multiple Paths

When resistors are connected across the same two points, providing multiple branches for the current to flow, they are in parallel. Imagine a river splitting into several smaller streams and then rejoining; the total flow of water is divided among the streams. This is exactly how current behaves in a parallel circuit.

The key characteristics of a parallel circuit are:

1. Same Voltage: The voltage drop is identical across every resistor in the parallel combination. V_total = V₁ = V₂ = V₃ = ...
2. Current Division: The total current supplied by the source (I_total) splits among the different branches. The sum of the currents in each branch equals the total current. I_total = I₁ + I₂ + I₃ + ...

{{VISUAL: diagram: A circuit diagram showing a battery connected to three resistors (R₁, R₂, R₃) in parallel branches. The current (I) from the battery is shown splitting into I₁, I₂, and I₃, which then recombine before returning to the battery.}}

Deriving the Formula for Equivalent Resistance

We use the same logic as before, but with the rules for parallel circuits.

1. We know I_total = I₁ + I₂ + I₃ + ...
2. Substitute I with V/R from Ohm's Law: V_total / R_eq = (V₁ / R₁) + (V₂ / R₂) + (V₃ / R₃) + ...
3. Since the voltage is the same across all branches (V_total = V₁ = V₂ = ...), we can cancel V from every term.

This gives us the formula for resistors in parallel.

{FORMULA: expr=1/Rₑ = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/Rₙ | symbols=Rₑ: Equivalent Resistance (Ω), R₁, R₂, ...: Individual Resistances (Ω)}

{KEY: type=warning | title=Critical Exam Tip: Don't Forget the Final Flip! | text=The most common mistake students make with the parallel formula is calculating the value of 1/Rₑ and forgetting to take the reciprocal at the end to find Rₑ. Always remember the last step: Rₑ = 1 / (answer). }

Special Case: Two Resistors in Parallel

For the very common case of only two resistors in parallel, we can simplify the formula algebraically: 1/Rₑ = 1/R₁ + 1/R₂ = (R₂ + R₁) / (R₁ × R₂) Flipping both sides gives us the "product over sum" rule, a massive time-saver in exams.

Rₑ (for two resistors) = (R₁ × R₂) / (R₁ + R₂)

A crucial insight from the parallel formula is that adding a resistor in parallel always decreases the total resistance. This is because you are adding another path for the current to flow, making it easier overall. The equivalent resistance in a parallel circuit is always smaller than the smallest individual resistance in the combination.

Solved Example 2: Basic Parallel Circuit (Easy)

Given: Two resistors R₁ = 6 Ω and R₂ = 3 Ω are connected in parallel to a 9 V battery.

To Find: (a) The total equivalent resistance (Rₑ). (b) The total current (I) drawn from the battery. (c) The current through each resistor (I₁ and I₂).

Solution:

1. Find the equivalent resistance (Rₑ). Since there are only two resistors, we can use the "product over sum" shortcut.

   Rₑ = (R₁ × R₂) / (R₁ + R₂)
   

   Rₑ = (6 Ω × 3 Ω) / (6 Ω + 3 Ω) = 18 / 9 = 2 Ω
   

   Notice that the result (2 Ω) is smaller than the smallest resistor (3 Ω).

2. Find the total current (I). Use Ohm's Law with the total voltage and equivalent resistance.

   I = V / Rₑ
   

   I = 9 V / 2 Ω = 4.5 A
   

3. Find the current through each resistor (I₁ and I₂). In a parallel circuit, the voltage across each resistor is the same as the source voltage (9 V). We apply Ohm's Law to each branch separately.

   • For R₁: I₁ = V / R₁ = 9 V / 6 Ω = 1.5 A
   • For R₂: I₂ = V / R₂ = 9 V / 3 Ω = 3.0 A

   Check: The sum of the branch currents should equal the total current. I₁ + I₂ = 1.5 A + 3.0 A = 4.5 A. This matches our result for the total current.

Final Answer: (a) The equivalent resistance is 2 Ω. (b) The total current is 4.5 A. (c) The current through the 6 Ω resistor is 1.5 A and through the 3 Ω resistor is 3.0 A.

Mixed (Series-Parallel) Circuits

Most real-world circuits are not purely series or purely parallel. They are combination or mixed circuits. The strategy to solve these is to break the circuit down into smaller, simpler parts.

The Strategy: Simplify and Redraw

1. Identify: Look for small, distinct groups of resistors that are either purely in series or purely in parallel.
2. Simplify: Calculate the equivalent resistance for one of these groups.
3. Redraw: Redraw the circuit, replacing the simplified group with its single equivalent resistor.
4. Repeat: Look at the new, simpler circuit and repeat the process until you are left with a single equivalent resistance for the entire circuit.

Solved Example 3: Mixed Circuit Analysis (Medium)

Given: A circuit with R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 12 Ω, and R₄ = 3 Ω connected to a 24 V source as shown in the description: R₁ is in series with the parallel combination of R₂ and R₃. This entire group is then in series with R₄. (This is a description of a standard mixed circuit diagram).

To Find: The total equivalent resistance (Rₑ) and the total current (I).

Solution:

1. Step 1: Simplify the parallel block. Resistors R₂ and R₃ are in parallel with each other. We calculate their equivalent resistance, let's call it R₂₃.

   R₂₃ = (R₂ × R₃) / (R₂ + R₃)
   

   R₂₃ = (6 Ω × 12 Ω) / (6 Ω + 12 Ω) = 72 / 18 = 4 Ω
   

2. Step 2: Redraw the circuit mentally. Now, our circuit simplifies to three resistors in series: R₁, R₂₃, and R₄. The circuit is effectively: (R₁) --- (R₂₃) --- (R₄)

3. Step 3: Calculate the final equivalent resistance. Since R₁, R₂₃, and R₄ are now in series, we simply add them up.

   Rₑ = R₁ + R₂₃ + R₄
   

   Rₑ = 4 Ω + 4 Ω + 3 Ω = 11 Ω
   

4. Step 4: Calculate the total current. Using Ohm's Law for the entire circuit.

   I = V / Rₑ
   

   I = 24 V / 11 Ω ≈ 2.18 A
   

Final Answer: The total equivalent resistance is 11 Ω and the total current is approximately 2.18 A.

Common Numerical Traps and Mistakes

Many students understand the concepts but lose marks due to simple calculation errors or formula mix-ups. Here are the most common traps to watch out for.

Exam-Style Multiple Choice Questions (MCQs)

Test your understanding with these questions modelled after competitive exams.

Question 1: Four resistors of 10 Ω each are connected in parallel. The equivalent resistance of the combination is: (a) 40 Ω (b) 10 Ω (c) 2.5 Ω (d) 0.4 Ω

💡 Solution: For n identical resistors R in parallel, the equivalent resistance is Rₑ = R / n. Here, R = 10 Ω and n = 4. Rₑ = 10 / 4 = 2.5 Ω. Correct Answer: (c). (Distractor (a) is the series result. Distractor (d) is 1/Rₑ without inverting)

Question 2: You are given three resistors of 2 Ω, 3 Ω, and 6 Ω. Which of the following total resistances is NOT possible to obtain by combining them? (a) 11 Ω (b) 1 Ω (c) 4.5 Ω (d) 5 Ω

💡 Solution: Let's test the options: (a) All in series: 2 + 3 + 6 = 11 Ω. Possible. (b) All in parallel: 1/Rₑ = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 6/6 = 1. So, Rₑ = 1 Ω. Possible. (c) (3 || 6) in series with 2: R_parallel = (3×6)/(3+6) = 18/9 = 2 Ω. R_total = 2 + 2 = 4 Ω. This is not 4.5 Ω. Let's try (2 || 3) in series with 6: R_parallel = (2×3)/(2+3) = 6/5 = 1.2 Ω. R_total = 1.2 + 6 = 7.2 Ω. Let's try (2 || 6) in series with 3: R_parallel = (2×6)/(2+6) = 12/8 = 1.5 Ω. R_total = 1.5 + 3 = 4.5 Ω. Possible. (d) 5 Ω: (2 in series with 3) parallel to 6: R_series = 2 + 3 = 5 Ω. R_total = (5×6)/(5+6) = 30/11 ≈ 2.7 Ω. Not 5. How about (3 in series with 6) parallel to 2? R_series = 9 Ω. R_total = (9×2)/(9+2) = 18/11 ≈ 1.6 Ω. How about (2 in series with 6) parallel to 3? R_series = 8 Ω. R_total = (8×3)/(8+3) = 24/11 ≈ 2.2 Ω. It seems 5 Ω is not possible with these combinations. Correct Answer: (d).

Question 3: In the circuit below, if another resistor is added in parallel to the 4 Ω resistor, the total current I drawn from the battery will: (a) Increase (b) Decrease (c) Remain the same (d) Become zero

💡 Solution: Adding a resistor in parallel always decreases the total equivalent resistance of the circuit. According to Ohm's Law (I = V / Rₑ), if the voltage V is constant and the total resistance Rₑ decreases, the total current I must increase. Correct Answer: (a).

Practice Problem Set

Solve these problems to master the topic. Final answers are provided for verification.

1. Find the equivalent resistance between points A and B for a circuit where a 5 Ω resistor and a 10 Ω resistor are in series, and this combination is in parallel with a 30 Ω resistor.

   💡 Answer: 10 Ω

2. Three resistors, 2 Ω, 4 Ω, and 6 Ω, are connected in parallel. This combination is connected in series with a 4 Ω resistor and a 12V battery. What is the current through the 6 Ω resistor?

   💡 Answer: 0.4 A

3. A wire of resistance 16 Ω is bent in the form of a circle. What is the effective resistance between the ends of any diameter?

   💡 Answer: 4 Ω

4. (Advanced) Find the equivalent resistance between points A and B of an infinite ladder of resistors, where each section consists of a 1 Ω resistor in series with a 2 Ω resistor in the rung.

   💡 Answer: 2 Ω

5. Five resistors are connected to form the sides of a cube. Find the equivalent resistance between two diagonally opposite corners. Assume each resistor is R.

   💡 Answer: (5/6)R

Quick Revision Cheatsheet

{FLASHCARD: q=What is the main difference between current and voltage behaviour in a series circuit? | a=In a series circuit, current is the same through all components, while voltage divides across them.} {FLASHCARD: q=What is the main difference between current and voltage behaviour in a parallel circuit? | a=In a parallel circuit, voltage is the same across all branches, while current divides among them.}

This final table summarises everything you need for quick revision.