Electrical Circuits I: Resistors in Series, Parallel, and Delta-Wye Connections

Resistors in Series

• Definition: Resistors are connected in series when they are chained end-to-end in a single path, so that the same electric current flows sequentially through every resistor in the connection.

• Voltage Characteristic:

  • The total voltage $v_T$ across a series combination of $n$ resistors equals the sum of the individual voltage drops across each resistor.   v_T = v_1 + v_2 + v_3 + v_4 + v_5 + v_6 + v_7 + \ndots + v_n

• Current Characteristic:

  • The current $i_T$ flowing through a series circuit is identical at all points and through every resistor.   i_T = i_1 = i_2 = i_3 = i_4 = i_5 = i_6 = i_7 = \ndots = i_n

• Total / Equivalent Resistance:

  • The total resistance $R_T$ of $n$ resistors in series is the sum of their individual resistance values.   R_T = R_1 + R_2 + R_3 + R_4 + R_5 + R_6 + R_7 + \ndots + R_n

• Application of Kirchhoff's Voltage Law (KVL):

  • Summing the potential differences around a closed series loop yields:   $v_s - v_1 - v_2 - v_3 - v_4 - v_5 - v_6 - v_7 = 0$

Resistors in Parallel

• Definition: Resistors are connected in parallel when both terminals of each resistor are connected across the same pair of common nodes, providing multiple alternate pathways for electric current.

• Voltage Characteristic:

  • The potential difference $v_T$ across every branch connected in parallel is identical.   v_T = v_1 = v_2 = v_3 = v_4 = \ndots = v_n

• Current Characteristic:

  • The total current $i_T$ entering a parallel combination equals the sum of the individual currents flowing through each parallel branch (by Kirchhoff's Current Law, KCL).   i_T = i_1 + i_2 + i_3 + i_4 + \ndots + i_n

• Total / Equivalent Resistance:

  • The reciprocal of the total equivalent resistance $R_T$ is equal to the sum of the reciprocals of the individual resistances.   $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \dots + \frac{1}{R_n}$

• Shortcut Formula for Two Resistors in Parallel:

  • When exactly two resistors $R_1$ and $R_2$ are connected in parallel, the total resistance is calculated as the product divided by the sum:   $R_T = \frac{R_1 \times R_2}{R_1 + R_2}$

Series-Parallel Circuit Analysis

Example 1: Finding Total Resistance ($R_{\text{TOTAL}}$) with a Voltage Source

• Given Circuit Parameters:

  • Source Voltage: $10\,\text{V}$

  • Resistors: $6\,\Omega$, $12\,\Omega$, $4\,\Omega$, $9\,\Omega$, $7\,\Omega$

• Step-by-Step Derivation:

  1. Combine the series branch on the right-hand side:      $R_A = 9\,\Omega + 7\,\Omega = 16\,\Omega$

  2. Combine $R_A$ in parallel with the $4\,\Omega$ resistor:      $R_B = \frac{4\,\Omega \times 16\,\Omega}{4\,\Omega + 16\,\Omega} = \frac{64}{20} = 3.2\,\Omega$

  3. Sum the remaining series resistors connected across terminals $a-b$:      $R_T = 6\,\Omega + 12\,\Omega + R_B = 6\,\Omega + 12\,\Omega + 3.2\,\Omega = 21.2\,\Omega$

• Final Answer:   $R_{\text{TOTAL}} = 21.2\,\Omega$

Example 2: Finding Total Resistance ($R_{\text{TOTAL}}$) with a Current Source

• Given Circuit Parameters:

  • Current Source: $3\,\text{mA}$

  • Resistors: $4\,\text{k}\Omega$, $10\,\text{k}\Omega$, $3\,\text{k}\Omega$, $5\,\text{k}\Omega$, $7\,\text{k}\Omega$

• Step-by-Step Derivation:

  1. Combine the rightmost loop series resistors:      $R_A = 3\,\text{k}\Omega + 5\,\text{k}\Omega + 7\,\text{k}\Omega = 15\,\text{k}\Omega$

  2. Combine $R_A$ in parallel with the $10\,\text{k}\Omega$ resistor:      $R_B = \frac{10\,\text{k}\Omega \times 15\,\text{k}\Omega}{10\,\text{k}\Omega + 15\,\text{k}\Omega} = \frac{150}{25} = 6\,\text{k}\Omega$

  3. Combine $R_B$ in series with the $4\,\text{k}\Omega$ resistor across terminals $a-b$:      $R_T = 4\,\text{k}\Omega + 6\,\text{k}\Omega = 10\,\text{k}\Omega$

• Final Answer:   $R_{\text{TOTAL}} = 10\,\text{k}\Omega$

Example 3: Finding Total Resistance ($R_{\text{TOTAL}}$) with a Millivolt Source

• Given Circuit Parameters:

  • Source Voltage: $200\,\text{mV}$

  • Resistors: $300\,\Omega$, $500\,\Omega$, $400\,\Omega$, $600\,\Omega$, $1.2\,\text{k}\Omega$

• Step-by-Step Derivation:

  1. Convert $1.2\,\text{k}\Omega$ to ohms: $1.2\,\text{k}\Omega = 1200\,\Omega$

  2. Combine the parallel pair of $600\,\Omega$ and $1200\,\Omega$:      $R_A = \frac{600\,\Omega \times 1200\,\Omega}{600\,\Omega + 1200\,\Omega} = \frac{720000}{1800} = 400\,\Omega$

  3. Add the series resistors along the single closed path:      $R_T = 300\,\Omega + R_A + 400\,\Omega + 500\,\Omega$      $R_T = 300\,\Omega + 400\,\Omega + 400\,\Omega + 500\,\Omega = 1600\,\Omega = 1.6\,\text{k}\Omega$

• Final Answer:   $R_{\text{TOTAL}} = 1600\,\Omega = 1.6\,\text{k}\Omega$

Example 4: Calculating Branch Currents ($i_s$, $i_1$, $i_2$)

• Given Circuit Parameters:

  • Voltage Source: $120\,\text{V}$

  • Resistors: $4\,\Omega$, $18\,\Omega$, $3\,\Omega$, $6\,\Omega$

• Step-by-Step Derivation:

  1. Find total resistance $R_T$:

  • Series combination of $3\,\Omega$ and $6\,\Omega$:        $R_A = 3\,\Omega + 6\,\Omega = 9\,\Omega$

  • Parallel combination of $18\,\Omega$ and $R_A = 9\,\Omega$:        $R_B = \frac{18\,\Omega \times 9\,\Omega}{18\,\Omega + 9\,\Omega} = \frac{162}{27} = 6\,\Omega$

  • Total equivalent resistance $R_T$:        $R_T = 4\,\Omega + R_B = 4\,\Omega + 6\,\Omega = 10\,\Omega$

  1. Calculate source current $i_s$:      $i_s = \frac{v_T}{R_T} = \frac{120\,\text{V}}{10\,\Omega} = 12\,\text{A}$

  2. Calculate current $i_1$ using KVL around the left loop:

  • Voltage drop across $4\,\Omega$ resistor: $v_4 = i_s \times 4\,\Omega = 12\,\text{A} \times 4\,\Omega = 48\,\text{V}$

  • Voltage across $18\,\Omega$ resistor: $v_{18} = 120\,\text{V} - 48\,\text{V} = 72\,\text{V}$

  • Current $i_1$:        $i_1 = \frac{v_{18}}{18\,\Omega} = \frac{72\,\text{V}}{18\,\Omega} = 4\,\text{A}$

  1. Calculate current $i_2$ using KCL at node $x$:      $i_s - i_1 - i_2 = 0 \implies i_2 = i_s - i_1 = 12\,\text{A} - 4\,\text{A} = 8\,\text{A}$

• Final Answers:

  • $(a)\,\,i_s = 12\,\text{A}$

  • $(b)\,\,i_1 = 4\,\text{A}$

  • $(c)\,\,i_2 = 8\,\text{A}$

Example 5: Finding Voltage, Delivered Power, and Dissipated Power

• Given Circuit Parameters:

  • Current Source: $5\,\text{A}$

  • Resistors: $30\,\Omega$, $7.2\,\Omega$, $64\,\Omega$, $6\,\Omega$, $10\,\Omega$

• Step-by-Step Derivation:

  1. Find total resistance $R_T$:

  • Series combination of $6\,\Omega$ and $10\,\Omega$:        $R_1 = 6\,\Omega + 10\,\Omega = 16\,\Omega$

  • Parallel combination of $16\,\Omega$ and $64\,\Omega$:        $R_2 = \frac{16\,\Omega \times 64\,\Omega}{16\,\Omega + 64\,\Omega} = \frac{1024}{80} = 12.8\,\Omega$

  • Series combination of $7.2\,\Omega$ and $12.8\,\Omega$:        $R_3 = 7.2\,\Omega + 12.8\,\Omega = 20\,\Omega$

  • Parallel combination of $30\,\Omega$ and $20\,\Omega$ across the current source:        $R_T = \frac{30\,\Omega \times 20\,\Omega}{30\,\Omega + 20\,\Omega} = \frac{600}{50} = 12\,\Omega$

  1. Calculate voltage $v$ across the current source:      $v = i_T \times R_T = 5\,\text{A} \times 12\,\Omega = 60\,\text{V}$

  2. Calculate power delivered by the current source:      $P = i_T \times v = 5\,\text{A} \times 60\,\text{V} = 300\,\text{W}$

  3. Calculate power dissipated in the $10\,\Omega$ resistor:

  • Current flowing through $30\,\Omega$ resistor: $i_A = \frac{60\,\text{V}}{30\,\Omega} = 2\,\text{A}$

  • Current entering the $7.2\,\Omega$ resistor branch: $i_B = 5\,\text{A} - 2\,\text{A} = 3\,\text{A}$

  • Voltage drop across $7.2\,\Omega$ resistor: $v_{7.2} = 3\,\text{A} \times 7.2\,\Omega = 21.6\,\text{V}$

  • Voltage across the parallel branch ($64\,\Omega$ and $16\,\Omega$ branch): $v_{\text{branch}} = 60\,\text{V} - 21.6\,\text{V} = 38.4\,\text{V}$

  • Current through the $16\,\Omega$ branch ($i_C$): $i_C = \frac{38.4\,\text{V}}{16\,\Omega} = 2.4\,\text{A}$

  • Power dissipated in the $10\,\Omega$ resistor:        $P_{10\,\Omega} = i_C^2 \times 10\,\Omega = (2.4\,\text{A})^2 \times 10\,\Omega = 5.76 \times 10 = 57.6\,\text{W}$

• Final Answers:

  • $(a)\,\,v = 60\,\text{V}$

  • $(b)\,\,P_{\text{delivered}} = 300\,\text{W}$

  • $(c)\,\,P_{10\,\Omega} = 57.6\,\text{W}$

Delta and Wye Connections

• Overview:

  • Certain resistor configurations cannot be simplified using standard series or parallel formulas alone. These networks are configured in Delta ($\Delta$) or Wye ($\text{Y}$) formations.

• Delta Connection ($\Delta$):

  • Consists of three resistors connected in a closed triangular loop between three terminals $a$, $b$, and $c$. It is also referred to as a $\Pi$ (Pi) network.

• Wye Connection ($\text{Y}$):

  • Consists of three resistors extending from a common central node to three outer terminals $a$, $b$, and $c$. It is also referred to as a $\text{T}$ (Tee) network.

• Transformation Relationships:

• Delta to Wye Transformation Formulas ($\Delta \rightarrow \text{Y}$):

  • Each resistor in the equivalent Wye network equals the product of the two adjacent Delta resistors divided by the sum of all three Delta resistors:   $R_1 = \frac{R_b \times R_c}{R_a + R_b + R_c}$   $R_2 = \frac{R_c \times R_a}{R_a + R_b + R_c}$   $R_3 = \frac{R_a \times R_b}{R_a + R_b + R_c}$

• Wye to Delta Transformation Formulas ($\text{Y} \rightarrow \Delta$):

  • Each resistor in the equivalent Delta network equals the sum of all possible pairwise products of Wye resistors divided by the opposite Wye resistor:   $R_a = \frac{R_1 \times R_2 + R_2 \times R_3 + R_3 \times R_1}{R_1}$   $R_b = \frac{R_1 \times R_2 + R_2 \times R_3 + R_3 \times R_1}{R_2}$   $R_c = \frac{R_1 \times R_2 + R_2 \times R_3 + R_3 \times R_1}{R_3}$

Delta-Wye Transformation Circuit Analysis

Example 6: Bridge Network with a $40\,\text{V}$ DC Supply

• Given Circuit Parameters:

  • DC Voltage Source: $40\,\text{V}$

  • Resistors: $5\,\Omega$ (series resistor), $100\,\Omega$, $125\,\Omega$, $25\,\Omega$, $40\,\Omega$, $37.5\,\Omega$

• Step-by-Step Derivation:

  1. Identify the top Delta network ($\Delta$) formed by $100\,\Omega$, $125\,\Omega$, and $25\,\Omega$ resistors:

  • Sum of Delta resistors: $R_{\Delta\text{ sum}} = 100\,\Omega + 125\,\Omega + 25\,\Omega = 250\,\Omega$

  1. Convert top Delta to equivalent Wye ($\text{Y}$) resistors $R_A$, $R_B$, $R_C$:      $R_A = \frac{100\,\Omega \times 125\,\Omega}{250\,\Omega} = \frac{12500}{250} = 50\,\Omega$      $R_B = \frac{100\,\Omega \times 25\,\Omega}{250\,\Omega} = \frac{2500}{250} = 10\,\Omega$      $R_C = \frac{125\,\Omega \times 25\,\Omega}{250\,\Omega} = \frac{3125}{250} = 12.5\,\Omega$

  2. Combine resulting series branches:

  • Left branch: $R_E = R_B + 40\,\Omega = 10\,\Omega + 40\,\Omega = 50\,\Omega$

  • Right branch: $R_D = R_C + 37.5\,\Omega = 12.5\,\Omega + 37.5\,\Omega = 50\,\Omega$

  1. Combine parallel branches $R_E$ and $R_D$:      $R_F = \frac{50\,\Omega \times 50\,\Omega}{50\,\Omega + 50\,\Omega} = 25\,\Omega$

  2. Calculate total equivalent resistance $R_T$:      $R_T = 5\,\Omega + R_A + R_F = 5\,\Omega + 50\,\Omega + 25\,\Omega = 80\,\Omega$

  3. Calculate total current supplied:      $I = \frac{V}{R_T} = \frac{40\,\text{V}}{80\,\Omega} = 0.5\,\text{A}$

  4. Calculate total power supplied:      $P = I \times V = 0.5\,\text{A} \times 40\,\text{V} = 20\,\text{W}$

• Final Answers:

  • Current supplied: $0.5\,\text{A}$

  • Power supplied: $20\,\text{W}$

Example 7: Finding Terminal Voltage $v$ Across a Current Source Network

• Given Circuit Parameters:

  • Current Source: $2\,\text{A}$

  • Resistors: $28\,\Omega$, $20\,\Omega$, $10\,\Omega$, $5\,\Omega$, $105\,\Omega$

• General Procedure:

  1. Identify Delta or Wye sub-networks within the circuit (such as the Delta formed by $28\,\Omega$, $20\,\Omega$, and $10\,\Omega$ or the Wye formed by $20\,\Omega$, $10\,\Omega$, and $5\,\Omega$).

  2. Apply the Delta-Wye or Wye-Delta conversion formulas to simplify the circuit into a single total resistance $R_T$ connected across the $2\,\text{A}$ source terminals.

  3. Solve for terminal voltage $v$ using Ohm's Law:      $v = 2\,\text{A} \times R_T$