Comprehensive Circuit Analysis Study Guide

Fundamental DC Circuit Rules and Node Analysis

• Node Labeling Procedure:

  • Node labeling is the mandatory first step when analyzing any electric circuit.
  • A node is defined as the point of intersection or connection boundary between two or more circuit elements.
  • Example Node Identification: In a single-loop series circuit connected to a DC power supply, Node 1 is the boundary between the power supply terminal and one end of resistor $R_1$; Node 2 is the connection point between $R_1$ and $R_2$; Node 3 is the connection point between $R_2$ and $R_3$.
  • Visual arrangement can be deceptive: elements that appear visually parallel in a schematic drawing may actually be connected sequentially in series if each element shares only one single node with an adjacent element.

• Series Circuit Properties:

  • Current Consistency: Components connected in series share the exact same nodes sequentially, meaning the same total current ($I_{\text{total}}$) flows continuously through every series element.
  • Total Resistance Formula:$R_{\text{total}} = R_1 + R_2 + R_3 + \dots + R_n$
  • Voltage Drop Across a Resistor (Ohm's Law):$V_{R_n} = I_{\text{total}} \times R_n$
  • Kirchhoff's Voltage Law (KVL) Constraint: The sum of all individual resistor voltage drops across a closed series loop must equal the source voltage ($V_{\text{source}}$). For a $10\,V$ source:         $V_{R_1} + V_{R_2} + V_{R_3} = 10\,V$

• Mandatory Engineering Metric Prefixes:

  • Prefix values are required knowledge for all calculations and are not provided on formula reference sheets:
    • Kilo ($k$) = $10^3$ ($1\,k\Omega = 1000\,\Omega$)
    • Milli ($m$) = $10^{-3}$ ($1\,mA = 1 \times 10^{-3}\,A$)

Parallel DC Circuits Analysis

• Parallel Circuit Rules and Node Structure:

  • In a purely parallel circuit, all parallel branches are connected directly between the same two primary nodes.
  • Branch Voltage: The voltage drop across every individual branch in a parallel circuit is constant and directly equals the total supply voltage ($V_{\text{branch}} = V_{\text{source}}$).

• Ohm's Law Applied to Individual Parallel Branches:

  • For a parallel circuit powered by a $20\,V$ DC source connected across three parallel resistors ($R_1 = 1\,k\Omega$, $R_2 = 2\,k\Omega$, $R_3 = 4\,k\Omega$):
    • Branch 1 Current ($I_1$):             $I_1 = \frac{V_{\text{source}}}{R_1} = \frac{20\,V}{1\,k\Omega} = 20\,mA$
    • Branch 2 Current ($I_2$):             $I_2 = \frac{V_{\text{source}}}{R_2} = \frac{20\,V}{2\,k\Omega} = 10\,mA$
    • Branch 3 Current ($I_3$):             $I_3 = \frac{V_{\text{source}}}{R_3} = \frac{20\,V}{4\,k\Omega} = 5\,mA$

• Total Parallel Current Calculation:

  • By Kirchhoff's Current Law (KCL), the total current entering the parallel network equals the sum of all individual branch currents:         $I_{\text{total}} = I_1 + I_2 + I_3$$I_{\text{total}} = 20\,mA + 10\,mA + 5\,mA = 35\,mA$

• Equivalent Parallel Resistance ($R_{\text{total}}$) Calculations:

  • Reciprocal Method using Least Common Multiple (LCM):$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$\frac{1}{R_{\text{total}}} = \frac{1}{1000} + \frac{1}{2000} + \frac{1}{4000}$
  • Finding the LCM ($4000$):         $\frac{1}{R_{\text{total}}} = \frac{4 + 2 + 1}{4000} = \frac{7}{4000}$$R_{\text{total}} = \frac{4000}{7}\,\Omega$
  • Decimal Precision Requirement: Convert fractions to standard decimal format rounded to exactly two decimal places:         $R_{\text{total}} \approx 571.43\,\Omega$

Series-Parallel Combination Circuit Analysis

• Circuit Parameters and Specifications:

  • Source Voltage ($V_s$) = $10\,V$
  • Resistor Values:
    • $R_1 = 1\,k\Omega$ ($1000\,\Omega$)
    • $R_2 = 500\,\Omega$
    • $R_3 = 2\,k\Omega$ ($2000\,\Omega$)
    • $R_4 = 1\,k\Omega$ ($1000\,\Omega$)
    • $R_5 = 2\,k\Omega$ ($2000\,\Omega$)
    • $R_6 = 2\,k\Omega$ ($2000\,\Omega$)

• Step 1: Complete Node Labeling:

  • Label all unique nodes in the circuit (5 total nodes):
    • Node 1: Positive terminal of $V_s$ to $R_1$.
    • Node 2: Junction of $R_1$, $R_3$, and $R_4$
    • Node 3: Junction of $R_4$, $R_5$, and $R_6$
    • Node 4: Common return rail connecting $R_3$, $R_5$, and $R_6$
    • Node 5: Return from $R_2$ to negative terminal of $V_s$

• Step 2: Circuit Reduction from Right-Hand Side (RHS):

  • Always reduce complex networks from the side furthest from the source (the right-hand side) back toward the power source to reduce the network to a single equivalent resistance across one voltage source.
  • First Reduction ($R_{\text{eq1}}$): Resistors $R_5$ and $R_6$ are in parallel between Nodes 3 and 4.         $R_{\text{eq1}} = R_5 \parallel R_6 = \frac{R_5 \times R_6}{R_5 + R_6}$$R_{\text{eq1}} = \frac{2000 \times 2000}{2000 + 2000} = \frac{4000000}{4000} = 1000\,\Omega = 1\,k\Omega$
  • Second Reduction ($R_{\text{eq2}}$): $R_{\text{eq1}}$ (located between Nodes 3 and 4) is in series with $R_4$.         $R_{\text{eq2}} = R_{\text{eq1}} + R_4$$R_{\text{eq2}} = 1\,k\Omega + 1\,k\Omega = 2\,k\Omega = 2000\,\Omega$
  • Third Reduction ($R_{\text{eq3}}$): $R_{\text{eq2}}$ (between Nodes 2 and 4) is in parallel with $R_3$ (between Nodes 2 and 4).         $R_{\text{eq3}} = R_3 \parallel R_{\text{eq2}} = \frac{2000 \times 2000}{2000 + 2000} = 1000\,\Omega = 1\,k\Omega$
  • Final Resistance ($R_{\text{total}}$): $R_1$, $R_{\text{eq3}}$, and $R_2$ are all connected in series across the supply terminals.         $R_{\text{total}} = R_1 + R_{\text{eq3}} + R_2$$R_{\text{total}} = 1\,k\Omega + 1\,k\Omega + 0.5\,k\Omega = 2.5\,k\Omega = 2500\,\Omega$

• Step 3: Calculating Total Source Current ($I_{\text{total}}$):$I_{\text{total}} = \frac{V_s}{R_{\text{total}}} = \frac{10\,V}{2500\,\Omega} = 0.004\,A = 4\,mA$

Voltage Drops and Branch Currents in Combination Circuits

• Tracing Total Current Flow:

  • By conventional current flow analysis, $I_{\text{total}}$ ($4\,mA$) leaves the source and passes entirely through $R_1$ before reaching Node 2.
  • After passing through the internal network, the total recombined current ($4\,mA$) passes entirely through $R_2$ before returning to the source.

• Outer Series Resistor Voltage Drops:

  • Voltage drop across $R_1$ ($V_{R1}$):         $V_{R1} = I_{\text{total}} \times R_1 = 4\,mA \times 1\,k\Omega = 4\,V$
  • Voltage drop across $R_2$ ($V_{R2}$):         $V_{R2} = I_{\text{total}} \times R_2 = 4\,mA \times 0.5\,k\Omega = 2\,V$

• Intermediate Loop Voltage Drop Calculation:

  • Applying Kirchhoff's Voltage Law to the main loop containing $V_s$, $R_1$, $R_3$, and $R_2$:         $V_s = V_{R1} + V_{R2} + V_{R3}$$10\,V = 4\,V + 2\,V + V_{R3}$$V_{R3} = 10\,V - 6\,V = 4\,V$
  • The voltage drop across $R_3$ ($4\,V$) acts as a secondary effective source voltage for all remaining parallel downstream network components connected between Nodes 2 and 4.

• Branch Currents and Downstream Drops:

  • Current through $R_3$ ($I_1$):         $I_1 = \frac{V_{R3}}{R_3} = \frac{4\,V}{2\,k\Omega} = 2\,mA$
  • Current entering the $R_4$ branch ($I_2$):         $I_2 = I_{\text{total}} - I_1 = 4\,mA - 2\,mA = 2\,mA$
  • Voltage drop across $R_4$ ($V_{R4}$):         $V_{R4} = I_2 \times R_4 = 2\,mA \times 1\,k\Omega = 2\,V$
  • Voltage drop across $R_5$ ($V_{R5}$):
    • Remaining voltage across parallel combination $R_5 \parallel R_6$ is equal to $V_{R3} - V_{R4} = 4\,V - 2\,V = 2\,V$
    • Therefore, $V_{R5} = 2\,V$

Wheatstone Bridge Circuits

• Circuit Drawing and Balance Conditions:

  • Students must know how to construct and draw a standard Wheatstone bridge circuit diagram.
  • A Wheatstone bridge is considered balanced when the differential output voltage across the central bridge terminals is zero ($V_{\text{out}} = 0\,V$).
  • Balance Ratio Formula:$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

• Application Procedures:

  • The formula for output voltage under unbalanced conditions is provided on the exam formula sheet.
  • Given one reference resistor value and the balance condition, solve for unknown branch resistances before substituting parameters into the output voltage equation.

AC Series RC Circuit Analysis

• Circuit Parameters Given in Problem Statement:

  • AC Source Voltage Amplitude ($V_s$) = $3\,V$
  • Frequency ($f$) = $1\,kHz = 1000\,Hz$
  • Resistance ($R$) = $1\,k\Omega = 1000\,\Omega$
  • Capacitance ($C$) = $10\,\mu F = 10 \times 10^{-6}\,F$

• **Step-by-Step