6.4 Series, Parallel & Combination DC Circuits

Series, Parallel & Combination DC Circuits

In electrotechnical engineering, electrical components and circuit loads can be interconnected in three fundamental configurations: series, parallel, and series-parallel combinations. Analyzing these circuits requires applying two universal physical principles formulated by German physicist Gustav Kirchhoff in 1845: Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL).

Mastering circuit reduction techniques is essential not only for passing electrical science examinations, but also for calculating fault currents, understanding ring final circuit continuity, predicting voltage drops across distribution runs, and diagnosing faults in complex control systems.

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Series Direct Current Circuits

A series circuit is an electrical configuration in which circuit components are connected end-to-end to form a single, continuous, unbranched loop between the power supply terminals.

Characteristics of Series Circuits

1. Uniform Current Throughout: Because there is only one conductive pathway, electric charge cannot accumulate at any point along the loop (conservation of electric charge). Therefore, the current flowing through every component in a series chain is identical: $I_T = I_1 = I_2 = I_3 = \dots = I_n$
2. Additive Total Resistance: The total opposition encountered by electric charge is the algebraic sum of the individual component resistances: $R_T = R_1 + R_2 + R_3 + \dots + R_n$
3. Summation of Voltage Drops (Kirchhoff's Voltage Law - KVL): Kirchhoff's Voltage Law states that the algebraic sum of all potential differences (voltages) around any closed conductive loop is equal to zero. In practical terms, the total electromotive force ($V_T$) supplied to a series loop equals the sum of the individual potential differences across each component: $V_T = V_1 + V_2 + V_3 + \dots + V_n$ Since $V_1 = I R_1$, $V_2 = I R_2$, and $V_3 = I R_3$, it follows that $V_T = I(R_1 + R_2 + R_3) = I R_T$.
4. The Voltage Divider Principle: The supply voltage divides across series components in direct proportion to their individual resistances. The voltage drop ($V_x$) across any specific resistor ($R_x$) in a series group is given by: $V_x = V_T \times \frac{R_x}{R_T}$

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Parallel Direct Current Circuits

A parallel circuit is an electrical configuration in which two or more components are connected across the same two common electrical nodes (or busbars), creating multiple independent parallel pathways for current.

Characteristics of Parallel Circuits

1. Identical Voltage Across All Branches: Because all branches are connected directly across the same common supply terminals, the potential difference across every parallel branch is identical to the total supply voltage: $V_T = V_1 = V_2 = V_3 = \dots = V_n$
2. Additive Branch Currents (Kirchhoff's Current Law - KCL): Kirchhoff's Current Law states that the algebraic sum of currents entering any electrical junction (node) must equal the algebraic sum of currents leaving that junction ($\sum I_{\text{in}} = \sum I_{\text{out}}$). Across a parallel bank, the total current drawn from the supply equals the sum of the individual branch currents: $I_T = I_1 + I_2 + I_3 + \dots + I_n$
3. Reciprocal Equivalent Resistance: Connecting additional parallel branches creates additional conductive pathways for electrons, which decreases the total equivalent circuit resistance. The reciprocal of the total resistance equals the sum of the reciprocals of the individual branch resistances: $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}$ $R_T = \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n} \right)^{-1}$

Special Case 1: Two Resistors in Parallel (The Product-over-Sum Rule)

When a parallel circuit comprises exactly two branches, the reciprocal formula simplifies to the widely used product-over-sum rule:

$R_T = \frac{R_1 \times R_2}{R_1 + R_2}$

Example: Two resistors of $20\ \Omega$ and $30\ \Omega$ connected in parallel yield: $R_T = \frac{20 \times 30}{20 + 30} = \frac{600}{50} = 12\ \Omega$ (Note: $12\ \Omega$ is strictly less than the smaller $20\ \Omega$ resistor.)

Special Case 2: $n$ Equal Resistors in Parallel

When $n$ identical resistors, each having a resistance of $R$, are connected in parallel, the equivalent resistance is simply the resistance of one resistor divided by the number of branches:

$R_T = \frac{R}{n}$

Example: Four identical $100\ \Omega$ lamps connected in parallel yield: $R_T = \frac{100}{4} = 25\ \Omega$

The Current Divider Principle (Two Branches)

In a two-branch parallel network, total current divides between the branches in inverse proportion to their resistances. The branch with the lower resistance carries the larger share of current:

$I_1 = I_T \times \frac{R_2}{R_1 + R_2}, \qquad I_2 = I_T \times \frac{R_1}{R_1 + R_2}$

Why BS 7671 Installation Wiring is Strictly Parallel

All standard electrical installations in buildings—including radial socket circuits, lighting circuits, and distribution boards—are wired in parallel. This design guarantees two critical operating conditions:

1. Every appliance and luminaire receives the full nominal supply voltage ($230\text{ V}$).
2. Each electrical device operates completely independently. Switching off or disconnecting one luminaire does not break the circuit or de-energize the other devices.

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Comparison: Series versus Parallel Circuits

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Systematic Reduction of Combination (Series-Parallel) Circuits

Most real-world electrotechnical networks are combination circuits, containing both series and parallel configurations. Solving these circuits requires a methodical step-by-step reduction technique:

1. Identify Isolated Clusters: Trace the circuit schematic from the source terminals to identify sub-groups that are purely in parallel or purely in series.
2. Reduce Parallel Branches: Replace each purely parallel cluster with a single equivalent resistor ($R_p$) using the product-over-sum rule or reciprocal formula.
3. Redraw the Simplified Circuit: Redraw the circuit replacing the parallel clusters with their equivalent single values.
4. Consolidate Series Elements: Add the equivalent parallel resistances to any remaining series resistors to determine the total circuit resistance ($R_T$).
5. Calculate Total Current: Apply Ohm's Law to calculate the total current drawn from the supply: $I_T = \frac{V_T}{R_T}$.
6. Work Backward to Determine Voltages and Branch Currents: Apply KVL to find voltage drops across series components ($V_s = I_T R_s$), determine the remaining voltage across parallel banks ($V_p = V_T - V_s$), and calculate individual branch currents using $I = \frac{V_p}{R_{\text{branch}}}$.

Comprehensive Worked Example: Combination Circuit Reduction

Problem: A $48\text{ V DC}$ power supply energizes a combination circuit. A series resistor $R_1 = 10\ \Omega$ is connected in series with a parallel cluster consisting of two resistors, $R_2 = 30\ \Omega$ and $R_3 = 60\ \Omega$. Calculate:

1. The equivalent resistance of the parallel group ($R_p$).
2. The total resistance of the entire circuit ($R_T$).
3. The total current supplied by the source ($I_T$).
4. The voltage drop across resistor $R_1$.
5. The voltage drop across the parallel cluster ($V_p$).
6. The branch currents flowing through $R_2$ and $R_3$.

Solution:

1. Calculate Equivalent Parallel Resistance ($R_p$): Using the product-over-sum rule for $R_2$ and $R_3$: $R_p = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{30 \times 60}{30 + 60} = \frac{1,800}{90} = 20\ \Omega$
2. Calculate Total Circuit Resistance ($R_T$): The series chain now consists of $R_1$ in series with $R_p$: $R_T = R_1 + R_p = 10\ \Omega + 20\ \Omega = 30\ \Omega$
3. Calculate Total Source Current ($I_T$): $I_T = \frac{V_T}{R_T} = \frac{48\text{ V}}{30\ \Omega} = 1.6\text{ A}$
4. Calculate Voltage Drop Across $R_1$ ($V_1$): Since all $1.6\text{ A}$ of total current passes through $R_1$: $V_1 = I_T \times R_1 = 1.6\text{ A} \times 10\ \Omega = 16.0\text{ V}$
5. Calculate Voltage Across the Parallel Cluster ($V_p$): Applying Kirchhoff's Voltage Law: $V_p = V_T - V_1 = 48.0\text{ V} - 16.0\text{ V} = 32.0\text{ V}$ (Verification via Ohm's Law: $V_p = I_T \times R_p = 1.6\text{ A} \times 20\ \Omega = 32.0\text{ V}$.)
6. Calculate Branch Currents ($I_2$ and $I_3$): Both branches experience the same $32.0\text{ V}$ potential difference: $I_2 = \frac{V_p}{R_2} = \frac{32.0\text{ V}}{30\ \Omega} \approx 1.067\text{ A}$ $I_3 = \frac{V_p}{R_3} = \frac{32.0\text{ V}}{60\ \Omega} \approx 0.533\text{ A}$ (Verification via Kirchhoff's Current Law: $I_2 + I_3 = 1.067 + 0.533 = 1.60\text{ A} = I_T$. KCL is verified.)

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Internal Resistance of Voltage Sources

In theoretical circuit analysis, a battery or DC generator is often treated as an "ideal voltage source" that maintains a perfectly constant terminal voltage regardless of how much current is drawn. In practical engineering, however, every chemical cell, battery, and generator has an internal physical resistance (symbolized by lowercase $r$).

Internal resistance arises from the electrical resistance of the chemical electrolyte, internal lead plates or electrodes, internal connection straps, and terminal posts.

Electromotive Force ($E$) versus Terminal Potential Difference ($V$)

• Electromotive Force ($E$): The total electrical energy per unit charge produced by chemical or mechanical action inside the source. It is measured in Volts and represents the open-circuit terminal voltage of the source when no current is flowing ($I = 0$).
• Terminal Potential Difference ($V$): The actual potential difference measured across the output terminals of the source when it is connected to a load and supplying current ($I > 0$).
• Lost Volts ($Ir$): As current $I$ leaves the battery, it must flow through the battery's own internal resistance $r$. By Ohm's Law, this current causes an internal voltage drop within the source: $V_{\text{lost}} = I \times r$

The Governing Source Formula

Applying Kirchhoff's Voltage Law around the circuit loop containing source $E$, internal resistance $r$, and external load $R_L$:

$E = V + I r = I R_L + I r = I (R_L + r)$

Transposed for the on-load terminal potential difference ($V$):

$V = E - I r$

Where:

• $V$ = On-load terminal potential difference (Volts)
• $E$ = Open-circuit electromotive force (Volts)
• $I$ = Load current (Amperes)
• $r$ = Source internal resistance (Ohms)

To calculate internal resistance from observed voltage sag:

$r = \frac{E - V}{I}$

Short-Circuit Current ($I_{sc}$)

If the output terminals of a voltage source are accidentally bridged by a zero-resistance short-circuit ($R_L \approx 0$):

$I_{sc} = \frac{E}{r}$

Because secondary cells (such as lead-acid or lithium-ion batteries) have very low internal resistance (often $0.01\ \Omega$ to $0.05\ \Omega$), their prospective short-circuit current can reach several thousand Amperes: $I_{sc} = \frac{12\text{ V}}{0.015\ \Omega} = 800\text{ A}$ This immense current causes catastrophic arc flash hazards, conductor melting, and potential battery explosion.

Worked Example: Source Internal Resistance

Problem: A lead-acid standby battery has an open-circuit electromotive force of $12.8\text{ V}$. When connected to an external load resistor of $2.3\ \Omega$, a current of $5.0\text{ A}$ flows through the circuit. Calculate:

1. The terminal potential difference ($V$) under this load.
2. The internal volt drop ("lost volts").
3. The internal resistance ($r$) of the battery.
4. The short-circuit current if the battery terminals are directly bridged.

Solution:

1. Terminal Potential Difference ($V$): $V = I \times R_L = 5.0\text{ A} \times 2.3\ \Omega = 11.5\text{ V}$
2. Internal Volt Drop ("Lost Volts"): $V_{\text{lost}} = E - V = 12.8\text{ V} - 11.5\text{ V} = 1.3\text{ V}$
3. Internal Resistance ($r$): $r = \frac{V_{\text{lost}}}{I} = \frac{1.3\text{ V}}{5.0\text{ A}} = 0.26\ \Omega$
4. Short-Circuit Current ($I_{sc}$): $I_{sc} = \frac{E}{r} = \frac{12.8\text{ V}}{0.26\ \Omega} \approx 49.23\text{ A}$