Analysis of Series and Parallel Resistive Circuits

Definition

A series and parallel resistive circuit is an electrical circuit in which two or more resistors are connected such that:

• In a series connection, resistors are connected end-to-end, so the same current flows through each resistor.
• In a parallel connection, resistors are connected across the same two points, so the same voltage appears across each resistor.

The analysis of such circuits involves finding the equivalent resistance and then using it to calculate current, voltage, and power in each part of the circuit.

Main Content

1. Series Resistive Circuits

• In a series circuit, resistors are connected one after another in a single path, so the current has only one route to follow.
• The same current flows through every resistor, while the total voltage is divided among the resistors according to their resistance values.

In a series circuit, the equivalent resistance is the sum of all individual resistances:

$$R_{eq} = R_1 + R_2 + R_3 + \cdots$$

This means the total resistance increases as more resistors are added in series. Since resistance is larger, the circuit current becomes smaller for a given supply voltage, according to Ohm’s law:

$$V = IR$$

Voltage division in series:
The source voltage is shared among the resistors. The voltage drop across each resistor is:

$$V_1 = IR_1,\quad V_2 = IR_2,\quad V_3 = IR_3$$

A resistor with a larger resistance gets a larger share of the total voltage.

Example:
If a 12 V source is connected to two resistors of 2 Ω and 4 Ω in series:

• Total resistance = 2 + 4 = 6 Ω
• Current = 12/6 = 2 A
• Voltage across 2 Ω resistor = 2 × 2 = 4 V
• Voltage across 4 Ω resistor = 2 × 4 = 8 V

So, the total voltage is divided as 4 V and 8 V, while the same 2 A current flows through both resistors.

2. Parallel Resistive Circuits

• In a parallel circuit, each resistor is connected across the same two nodes, creating multiple paths for current.
• The voltage across every resistor in parallel is the same, but the current splits according to the resistance of each branch.

The equivalent resistance of parallel resistors is found using:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$$

For two resistors, this can also be written as:

$$R_{eq} = \frac{R_1R_2}{R_1 + R_2}$$

The equivalent resistance of a parallel combination is always less than the smallest individual resistor because the current has more than one path.

Current division in parallel:
The total current from the source divides among the branches:

$$I = I_1 + I_2 + I_3$$

The branch current through each resistor is:

$$I_1 = \frac{V}{R_1},\quad I_2 = \frac{V}{R_2},\quad I_3 = \frac{V}{R_3}$$

Thus, lower resistance branches carry more current.

Example:
If a 12 V source is connected to two resistors of 6 Ω and 3 Ω in parallel:

• Voltage across each resistor = 12 V
• Current through 6 Ω resistor = 12/6 = 2 A
• Current through 3 Ω resistor = 12/3 = 4 A
• Total current = 2 + 4 = 6 A
• Equivalent resistance = 12/6 = 2 Ω

This shows that the total resistance is much smaller than either branch resistance.

3. Combination of Series and Parallel Networks

• Many real circuits are not purely series or purely parallel, but a combination of both.
• Such networks are simplified step by step by replacing groups of resistors with their equivalent resistance until a single resistance remains.

The analysis of combined circuits depends on identifying which resistors are in series and which are in parallel:

• Resistors in series carry the same current.
• Resistors in parallel have the same voltage.

Method of simplification:

1. Identify the simplest series or parallel group.
2. Replace it with its equivalent resistance.
3. Repeat the process until the circuit becomes a single equivalent circuit.
4. Use Ohm’s law to find total current.
5. Work backward to find individual branch currents and voltage drops.

Example:
Suppose a 4 Ω resistor is in series with a parallel combination of 6 Ω and 3 Ω.

• First, find the parallel equivalent: $$R_p = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 \, \Omega$$

• Then add the series resistor: $$R_{eq} = 4 + 2 = 6 \, \Omega$$

• If the supply is 12 V, then total current: $$I = \frac{12}{6} = 2 \, A$$

• Voltage across the 4 Ω resistor = 2 × 4 = 8 V

• Voltage across the parallel branch = 12 - 8 = 4 V
• Current in 6 Ω branch = 4/6 = 0.67 A
• Current in 3 Ω branch = 4/3 = 1.33 A

This shows how series-parallel analysis helps in determining all circuit quantities accurately.

Working / Process

1. Identify the circuit type and resistor arrangement

Examine the circuit carefully to determine which resistors are connected in series, which are in parallel, and whether the circuit is a combination network. This step is important because incorrect identification leads to wrong calculations.

2. Find the equivalent resistance

Use the series formula $R_{eq} = R_1 + R_2 + \cdots$ for series parts and the parallel formula $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$ for parallel parts. Simplify the circuit gradually until only one equivalent resistance remains.

3. Calculate current, voltage, and power

Apply Ohm’s law to find total current from the source. Then use voltage division in series circuits and current division in parallel circuits to determine individual resistor values. If required, calculate power using: $$P = VI,\quad P = I^2R,\quad P = \frac{V^2}{R}$$ Finally, verify results using Kirchhoff’s laws to ensure the circuit is analyzed correctly.

Advantages / Applications

• Helps in determining total resistance, current distribution, and voltage drops in electrical networks accurately.
• Forms the basis for analyzing and designing practical circuits such as lighting systems, resistor banks, battery loads, and electronic biasing networks.
• Useful in troubleshooting faults, estimating power loss, and selecting appropriate resistor values for safe and efficient circuit operation.

Summary

• Series circuits have one current path and the same current flows through all resistors.
• Parallel circuits provide multiple current paths and the same voltage appears across each branch.
• Combined circuits are solved by simplifying step by step using equivalent resistance and applying Ohm’s law.