Combination circuits Knowledge-Based Study

Study notes

1. What is a Combination Circuit?

   In direct-current (DC) networks, practical circuits rarely consist of purely series or purely parallel arrangements. A combination circuit (also called a series-parallel circuit) contains components connected in both configurations simultaneously.

   To analyze these networks effectively, you must recall the governing rules for each basic connection type:

   • Series connections: Components are connected end-to-end along a single conducting path. The electric current $I$ is identical through all series components: $$I_{\text{total}} = I_1 = I_2 = \dots = I_n$$ The total potential difference across the group is the sum of individual voltage drops: $$V_{\text{total}} = V_1 + V_2 + \dots + V_n$$ Equivalent resistance: $$R_{\text{eq}} = R_1 + R_2 + \dots + R_n$$
   • Parallel connections: Components are connected across the same common electrical nodes. The potential difference $V$ across each parallel branch is identical: $$V_{\text{total}} = V_1 = V_2 = \dots = V_n$$ The total current entering the junction equals the sum of branch currents: $$I_{\text{total}} = I_1 + I_2 + \dots + I_n$$ Equivalent resistance: $$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$$

   Step-by-Step Circuit Reduction

   The primary method for simplifying a combination circuit is the reduction method. This involves identifying local sets of resistors that are strictly in series or strictly in parallel, replacing them with their equivalent resistance, and repeating the process until a single total equivalent resistance remains.

   Systematic Reduction Procedure:

   1. Identify common nodes: Trace paths between junctions. Two resistors are in parallel only if both pairs of their respective terminals share the same electrical nodes. Two resistors are in series only if the current leaving one must entirely enter the other with no intermediate junction.
   2. Start furthest from the source: Locate the innermost sub-circuits that form clear parallel or series pairs and calculate their equivalent values.
   3. Redraw the circuit: Replace each identified combination with an equivalent resistor $R_{\text{eq, sub}}$. Redrawing prevents errors in complex networks.
   4. Repeat until simplified: Continue combining adjacent series and parallel elements until only one equivalent resistance, $R_{\text{total}}$, is connected across the voltage source.

   Application: Resistance Reduction

   Consider a circuit powered by an ideal DC source of $V = 24\text{ V}$. Resistor $R_1 = 4\ \Omega$ is connected in series with a parallel bank containing two resistors, $R_2 = 6\ \Omega$ and $R_3 = 12\ \Omega$.

   Step 1: Simplify the Parallel Sub-circuit

   Resistors $R_2$ and $R_3$ share the same pair of nodes, meaning they are connected strictly in parallel. Their equivalent resistance $R_{23}$ is determined by:

   $$\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{6\ \Omega} + \frac{1}{12\ \Omega} = \frac{2}{12\ \Omega} + \frac{1}{12\ \Omega} = \frac{3}{12\ \Omega}$$$$R_{23} = \frac{12}{3} = 4\ \Omega$$

   Step 2: Combine with the Series Resistor

   The entire current from the source must pass through $R_1$ before reaching the parallel junction containing $R_{23}$. Therefore, $R_1$ and $R_{23}$ are in series:

   $$R_{\text{total}} = R_1 + R_{23} = 4\ \Omega + 4\ \Omega = 8\ \Omega$$

   The total equivalent resistance of the entire combination circuit is $8\ \Omega$.

   Determining Branch Currents and Potential Differences

   Once $R_{\text{total}}$ is determined, circuit analysis proceeds in reverse (expanding back out from the single equivalent resistor to the original network) using Ohm's law and Kirchhoff's laws.

   Continuing the Worked Example:

   With an applied electromotive force of $V = 24\text{ V}$ and $R_{\text{total}} = 8\ \Omega$:

   • Total circuit current: $$I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{24\text{ V}}{8\ \Omega} = 3\text{ A}$$
   • Voltage drop across $R_1$: Because all current must pass through $R_1$, $I_1 = I_{\text{total}} = 3\text{ A}$. $$V_1 = I_1 \cdot R_1 = 3\text{ A} \cdot 4\ \Omega = 12\text{ V}$$
   • Voltage across the parallel bank: According to Kirchhoff's Voltage Law, the remaining potential difference drops across the parallel pair: $$V_{\text{parallel}} = V - V_1 = 24\text{ V} - 12\text{ V} = 12\text{ V}$$ Both $R_2$ and $R_3$ experience this full $12\text{ V}$ drop: $V_2 = V_3 = 12\text{ V}$.
   • Branch currents through $R_2$ and $R_3$: $$I_2 = \frac{V_2}{R_2} = \frac{12\text{ V}}{6\ \Omega} = 2\text{ A}$$ $$I_3 = \frac{V_3}{R_3} = \frac{12\text{ V}}{12\ \Omega} = 1\text{ A}$$

   Notice that $I_2 + I_3 = 2\text{ A} + 1\text{ A} = 3\text{ A} = I_{\text{total}}$, which satisfies Kirchhoff's Current Law at the junction.

   Summary and Analytical Checklist

   • Never assume components are purely series or parallel at a glance: Verify connections by tracking nodes and junctions. If current branches between two resistors, they are not in series. If resistors do not share both endpoints, they are not in parallel.
   • Work from the inside out: Simplify the most deeply nested parallel or series groups first, then substitute their single equivalent value.
   • Check with conservation laws: The sum of powers dissipated in all individual resistors must equal the total power delivered by the source: $$P_{\text{total}} = V \cdot I_{\text{total}} = \sum I_k^2 R_k$$