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Annex 1 (BE01) — Lesson BE.01.08 • TLO: 28XX-MAIN-1002 • ELOs: 28XX-MAIN-1002ba, 1002be, 1002bh, 1002bl

Series-Parallel Resistive Circuits

A series-parallel (combination) circuit combines series elements and parallel banks within the same network. To the DC power supply, the entire combination network still appears as a single total resistance ( $R_T$ ). Solving or troubleshooting any series-parallel circuit requires applying series rules to the series sections and parallel rules to the parallel sections in a strict six-step order of operations.

1. The Mandatory 6-Step Reduction & Solution Process

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VA R1 (Series) R2 R3 (RL) 1. Original Circuit (R1 + [R2 ∥ R3]) Step 1–2 VA R1 REQ 2. Redrawn Series Circuit (RT = R1 + REQ)

Step-by-Step Order of Operations (`ELO 28XX-MAIN-1002ba`)

Step	Action	Governing Rule & Formula
Step 1	Work Out All Parallel Sections ( $R_{EQ}$ )	Reduce every parallel bank to a single Equivalent Resistance ( $R_{EQ}$ ) using the Reciprocal (`RSOR`), Product-Over-Sum ( $\frac{R_2 \times R_3}{R_2 + R_3}$ ), or Value-Over-Number ( $\frac{R}{N}$ ) method.
Step 2	Redraw the Circuit & Solve for Total Resistance ( $R_T$ )	Redraw the schematic as a simple series circuit with $R_{EQ}$ replacing the parallel bank, then add all series resistances together: $R_T = R_1 + R_{EQ}$
Step 3	Find Total Current ( $I_T$ )	Using total resistance ( $R_T$ ) and applied voltage ( $V_A$ ), calculate total circuit current via Ohm’s Law: $I_T = \frac{V_A}{R_T}$ Because $R_1$ and the parallel bank ( $R_{EQ}$ ) are in series with the source, $I_{R1} = I_{REQ} = I_T$ .
Step 4	Solve for Series Voltage Drops ( $V_{R1}, V_{REQ}$ )	Using common series current ( $I_T$ ), calculate the voltage drop across each series element and across the equivalent parallel bank: $V_{R1} = I_T \times R_1 \qquad\text{and}\qquad V_{REQ} = I_T \times R_{EQ}$ (Verify with KVL: $V_A = V_{R1} + V_{REQ}$ .)
Step 5	Solve for Parallel Branch Currents ( $I_{R2}, I_{R3}$ )	Expand $R_{EQ}$ back into its parallel branches. Because voltage is common in parallel, the voltage across every branch in that bank equals $V_{REQ}$ ( $V_{R2} = V_{R3} = V_{REQ}$ ). Calculate each branch current: $I_{R2} = \frac{V_{REQ}}{R_2} \qquad\text{and}\qquad I_{R3} = \frac{V_{REQ}}{R_3}$ (Verify with KCL: $I_{R2} + I_{R3} = I_T$ .)
Step 6	Solve for Individual & Total Power ( $P$ )	Calculate power for each resistor ( $P_x = I_x \times V_x$ ) and verify that the sum of all individual resistor powers equals total circuit power: $P_T = P_{R1} + P_{R2} + P_{R3} = I_T \times V_A$

2. Loaded Voltage Dividers & The 10:1 Rule

When an external device or circuit (Load, $R_L$ ) is connected to tap a specific voltage off a series voltage divider, $R_L$ is wired in parallel with one of the voltage divider’s resistors ( $R_2$ ), transforming the circuit into a series-parallel loaded voltage divider.

Key Loaded Voltage Divider Concepts

The Load ( $R_L$ ): Represents the external component or subsystem being powered by the voltage divider (absorbs power to perform work).
The Loading Effect: Connecting $R_L$ $R_{L}$ in parallel with $R_2$ $R_{2}$ creates a parallel bank whose equivalent resistance ( $R_{EQ} = R_2 \parallel R_L$ $R_{E Q} = R_{2} ∥ R_{L}$ ) is always smaller than $R_2$ alone.
- Because $R_{EQ}$ decreases, total circuit resistance ( $R_T = R_1 + R_{EQ}$ ) decreases, total current ( $I_T$ ) increases, the voltage drop across series resistor $R_1$ ( $V_{R1} = I_T \times R_1$ ) increases, and the output voltage across the load ( $V_L$ / $V_{\text{out}}$ ) DECREASES (“droops”).
- Proportionality Summary: $\text{Smaller } R_L \implies \text{Smaller } R_{EQ} \implies \text{Smaller Output Voltage } (V_L) \implies \text{Larger Load Current } (I_L)$
The 10:1 Loading Rule: To prevent unacceptable output voltage loss when the load is connected, a voltage divider is designed so that the load resistance ( $R_L$ ) is at least $10\text{ times}$ greater than the bleeder resistor ( $R_2$ ) ( $R_L \ge 10 \times R_2$ $R_{L} \geq 10 \times R_{2}$ ).
- Example: If $R_1 = 2,\text{k}\Omega$ and $R_2 = 4,\text{k}\Omega$ , choosing $R_L \ge 40,\text{k}\Omega$ keeps $R_{EQ}$ ( $4,\text{k}\Omega \parallel 40,\text{k}\Omega = 3.64,\text{k}\Omega$ ) close to $4,\text{k}\Omega$ , minimizing voltage droop.
Load Current ( $I_L$ ) vs. Bleeder Current ( $I_{R2}$ ):
- Load Current ( $I_L$ ): Current flowing through the load device ( $R_L$ ).
- Bleeder Current ( $I_{R2}$ ): Current flowing through the divider resistor ( $R_2$ ) in the branch parallel to the load.

3. Worked Circuit & Fault Demonstrations

Demonstration 1: Complete 6-Step Calculation

Given: $V_A = 15,\text{V}$ , series resistor $R_1 = 1,\text{k}\Omega$ , and parallel branches $R_2 = 2,\text{k}\Omega \parallel R_3 = 3,\text{k}\Omega$ .

Step 1 ( $R_{EQ}$ ): $R_{EQ} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{2,\text{k}\Omega \times 3,\text{k}\Omega}{2,\text{k}\Omega + 3,\text{k}\Omega} = \frac{6\times 10^6}{5,\text{k}\Omega} = \mathbf{1.20,\text{k}\Omega}$
Step 2 ( $R_T$ ): $R_T = R_1 + R_{EQ} = 1,\text{k}\Omega + 1.2,\text{k}\Omega = \mathbf{2.20,\text{k}\Omega}$
Step 3 ( $I_T$ ): $I_T = \frac{V_A}{R_T} = \frac{15,\text{V}}{2.2,\text{k}\Omega} = \mathbf{6.82,\text{mA}}$ ( $I_{R1} = I_{REQ} = 6.82,\text{mA}$ )
Step 4 (Series Drops):
- $V_{R1} = I_T \times R_1 = 6.818,\text{mA} \times 1,\text{k}\Omega = \mathbf{6.82,\text{V}}$
- $V_{REQ} = I_T \times R_{EQ} = 6.818,\text{mA} \times 1.2,\text{k}\Omega = \mathbf{8.18,\text{V}}$ ( $= V_{R2} = V_{R3}$ )
Step 5 (Branch Currents):
- $I_{R2} = \frac{V_{R2}}{R_2} = \frac{8.182,\text{V}}{2,\text{k}\Omega} = \mathbf{4.09,\text{mA}}$
- $I_{R3} = \frac{V_{R3}}{R_3} = \frac{8.182,\text{V}}{3,\text{k}\Omega} = \mathbf{2.73,\text{mA}}$ (KCL Check: $4.09 + 2.73 = 6.82,\text{mA} = I_T$ )
Step 6 (Power):
- $P_{R1} = \mathbf{46.49,\text{mW}}$ , $P_{R2} = \mathbf{33.47,\text{mW}}$ , $P_{R3} = \mathbf{22.31,\text{mW}}$ , $P_T = \mathbf{102.27,\text{mW}}$ .

Demonstration 2: Faulted Circuit — SHORT in a Parallel Branch (`PA #2`)

Given: $V_A = 12,\text{V}$ , series $R_1 = 470,\Omega$ , $R_2 = \text{SHORT}$ ( $0,\Omega$ ), $R_3 = 2.4,\text{k}\Omega$ .

Analysis: Because $R_2$ $R_{2}$ is shorted ( $0,\Omega$ $0 Ω$ ), it shorts out the entire parallel bank:
- $R_{EQ} = \mathbf{0,\Omega\text{ (MIN)}}$ and $V_{REQ} = V_{R2} = V_{R3} = \mathbf{0,\text{V (MIN)}}$ .
- Total circuit resistance reduces to only the series resistor: $R_T = R_1 = \mathbf{470,\Omega}$ .
- Full applied voltage drops across $R_1$ : $V_{R1} = V_A = \mathbf{12,\text{V}}$ .
- Total current increases to $I_T = I_{R1} = \frac{12,\text{V}}{470,\Omega} = \mathbf{25.53,\text{mA}}$ .
- $100%$ of $I_T$ flows through the shorted branch ( $I_{R2} = \mathbf{25.53,\text{mA [MAX]}}$ ), while intact branch $R_3$ is starved of current ( $I_{R3} = \mathbf{0,\text{A [MIN]}}$ ).

Demonstration 3: Faulted Circuit — OPEN in a Parallel Branch (`PA #3`)

Given: $V_A = 1.5,\text{V}$ , series $R_1 = 1,\Omega$ , parallel $R_2 = 4,\Omega$ , $R_3 = \text{OPEN}$ ( $\infty,\Omega$ ).

Analysis: With branch $R_3$ $R_{3}$ open, no current can flow through $R_3$ $R_{3}$ ( $I_{R3} = \mathbf{0,\text{A [MIN]}}$ $I_{R 3} = 0 A [MIN]$ ), so the circuit collapses into a simple series circuit of $R_1$ $R_{1}$ and $R_2$ $R_{2}$ :
- $R_{EQ} = R_2 = \mathbf{4,\Omega}$ (increased because a parallel branch was lost!).
- $R_T = R_1 + R_2 = 1,\Omega + 4,\Omega = \mathbf{5,\Omega}$ .
- $I_T = I_{R1} = I_{R2} = \frac{1.5,\text{V}}{5,\Omega} = \mathbf{300,\text{mA}}$ .
- $V_{R1} = 300,\text{mA} \times 1,\Omega = \mathbf{300,\text{mV}}$ ; $V_{REQ} = V_{R2} = V_{R3} = \mathbf{1.20,\text{V}}$ (Note: the open terminals of $R_3$ still measure the $1.20,\text{V}$ parallel bank voltage!).

4. Bench & Troubleshooting Application (`BE.01.08` Lab)

In the BE.01.08 breadboard lab, the standard reference circuit across $V_A = 5,\text{VDC}$ uses:

Series Resistor: $R_1 = 330,\Omega$
Parallel Bank: $R_2 = 1.2,\text{k}\Omega \parallel R_3 = 2.7,\text{k}\Omega$
Nominal Values: $R_{EQ} = 830.77,\Omega$ , $R_T = 1.16,\text{k}\Omega$ , $I_T = 4.31,\text{mA}$ (measured $\approx 4.12,\text{mA}$ under slight source drop), $V_{R1} \approx 1.36\text{–}1.42,\text{V}$ , $V_{R2} = V_{R3} = V_{REQ} \approx 3.35\text{–}3.58,\text{V}$ .

Diagnostic Decision Matrix for Series-Parallel Faults

Component Fault	Total Resistance ( $R_T$ )	Total Current ( $I_T$ )	Series Drop ( $V_{R1}$ )	Parallel Bank Voltage ( $V_{REQ} = V_{R2} = V_{R3}$ )	Key Identifying Signature
$R_1$ (Series) OPEN	`OL` ( $\infty$ )	$0,\text{A}$	$V_A$ ( $5.0,\text{V}$ )	$0.0,\text{V}$	Full source voltage across $R_1$ ; $0,\text{V}$ across parallel bank.
$R_1$ (Series) SHORTED	$R_{EQ}$ ( $830,\Omega$ ) $\downarrow$	Increases ( $\uparrow$ )	$0.0,\text{V}$	$V_A$ ( $5.0,\text{V}$ ) $\uparrow$	$0,\text{V}$ across $R_1$ ; full source voltage appears across the parallel bank.
$R_2$ or $R_3$ (Parallel) OPEN	Increases ( $\uparrow$ ) ( $R_1 + R_{\text{intact}}$ )	Decreases ( $\downarrow$ )	Decreases ( $\downarrow$ )	Increases ( $\uparrow$ )	Lower $I_T$ causes $R_1$ to drop less voltage, so by KVL ( $V_{REQ} = V_A - V_{R1}\downarrow$ ), parallel bank voltage increases!
$R_2$ or $R_3$ (Parallel) SHORTED	$R_1$ ( $330,\Omega$ ) $\downarrow$	Increases ( $\uparrow$ max)	$V_A$ ( $5.0,\text{V}$ ) $\uparrow$	$0.0,\text{V}$	Shorted branch zeroes out $R_{EQ}$ and $V_{REQ}$ ; $R_1$ drops $100%$ of $V_A$ .
$R_2$ or $R_3$ Value INCREASED	Increases ( $\uparrow$ )	Decreases ( $\downarrow$ )	Decreases ( $\downarrow$ )	Increases ( $\uparrow$ )	Measure isolated branch resistances with power off to pinpoint which resistor drifted high.
$R_2$ or $R_3$ Value DECREASED	Decreases ( $\downarrow$ )	Increases ( $\uparrow$ )	Increases ( $\uparrow$ )	Decreases ( $\downarrow$ )	Lower $R_{EQ}$ pulls $V_{REQ}$ down and drives $V_{R1}$ up.

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