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Circuit Analysis

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Connor

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Section 1: Basic Circuit Analysis Groundwork

NameSymbolUseUnits
AmpACurrent FlowA
ColumbCElectric ChargeA · s
VoltVElectric Potentialkg · m² · s⁻³ · A⁻¹
OhmΩResistancekg · m² · s⁻³ · A⁻²
WattWPowerkg · m² · s⁻³
JouleJEnergykg · m² · s⁻²
FaradFCapacitancekg⁻¹ · m⁻² · s⁴ · A²
HenryHInductancekg · m² · s⁻² · A⁻²
HertzHzFrequencys⁻¹

Basic Equations

The Fundamental Equations we will use often throughout this course are.

V=I∗RV=I*R

V=IR is the simplest and most fundamental equation. Memorize it, we will be using it alot. Voltage(V) is Equal to Current(I) Multiplied by Resistance(R). You can rearrange this equation as needed. To solve for Current and Resistance.

I=R/VI=R/V
R=V/IR=V/I

Symbols

Section 2: Parallel and Series Calculations

The equation for Parallel Resistance is

1/R1+1/R2+1/Rn...=1/R1/R_1+1/R_2+1/R_n…=1/R

The Equation for Series Resistance is

R1+R2+Rn...=RR_1+R_2+R_n… =R

Exercise 1:

Section 3: KVL and KCL

Kirchhoff’s Voltage Law: All voltages in a Loop must add up to Zero.

EXAMPLE IMAGE

MATHMATH

Kirchhoff’s Current Law: All Currents entering a node must add up to Zero

EXAMPLE IMAGE

MATHMATH

Exercise 1: Solve for the Voltage in the Resistor R2

Exercise 2: Solve for the Current in Resistor R2

Voltage Drop

A necessary component of using KVL Effectively is Voltage Drop. Given two Resistors in Series, the Voltage drop across each of them will be proportional to the resistance. As a example, a 12v source, that runs through two resistors R1=1ohm and R2=2ohm. The Voltage Drop will be equal to.

VR1=VSource∗R1R1+R2V_{R1}=V_{Source}*\frac{R1}{R1+R2}

This is true for any number of resistors in Series. The denominator represents Total Resistance in Series.

Rember this, it will be important later Voltage Division can be used to calculate the Voltage Drop across any Resistor. The Drop may not be relative to your ground(I will explain this with the Node Current method), but it will tell you the drop of the voltage across a resistor which can be used to calculate current.

Section 4: Power Equations

You can solve many equations using Conservation of Energy. Fundamentally this involved Power equations, the sum of all the power entering or leaving the system must be Zero.

The fundamental equations are.

Power is Equal to Current(I) times Voltage(V)

P=I∗VP=I*V

Power is equal to Current(I) Squared times Resistance

P=I2∗RP=I^2*R

Power is equal to Voltage(V) Squared times Resistance

P=V2/RP=V^2/R

Exercise 1:

Section 5: Mesh Voltage and Node Current Methods

Two methods, that fundamentally are extensions of KVL/KCL are the Mesh Voltage and Node Current Methods. I will describe them together, then go into them separately. The Mesh Voltage and Node Current Methods, involve getting a series of Linear Equations, then solving them simultaneously. You can solve these linear equations manually, but i highly suggest using a calculator for this.

Mesh Voltage

Mesh Voltage fundamentally relies on Kirchoff’s Voltage Law, the Voltage in a Loop must equal Zero.

Node Current

Section 6: Source Conversion and Voltage Dividers

When evaluating a circuit, it may be convenient or necessary to convert between a voltage source and a Current source.

Section 7: Op Amps

Previously, before Digital Electronics Opamps were used to do analog Math operations.

Op Amps have a couple features and properties that make them very useful. Understanding these features and properties makes solving simple circuits with them extremely straight forward.

The Op Amp will try to make the Voltage of both Inputs equal. A Ideal Op Amp will not allow any current to pass through either input. The Voltage provided by the Output cannot exceed the Supply(+/-).

Features and Equations.

  • Input(+)
    Non-Inverting Input, Vin+
    Must Equal Vin-
    Vin- = Vin+
  • Input(-)
    Inverting Input, Vin-
    Must Equal Vin+
    Vin+ = Vin-
  • Vout
    Output
    Vout =A(Vin+ – Vin-)
  • Supply(+)
    Provides power to the Op Amp
    Vout Cannot exceed Supply(+)
  • Supply(-)
    Provides power to the Op Amp
    Vout Cannot exceed Supply(-)

Amplitude

Integrator Circuit

Sumer Circuit

Example 1:

Section 8: Applications of Calculus and Inductors/Capacitors

Capacitors and Inductors store energy in various ways. Capacitors store energy in Electric Potential, Inductors store Energy in a magnetic field. In DC circuits, when you apply a voltage and current to one of them, they eventually become Saturated.

The Natural Response, is when a Saturated Capacitor or Inductor is suddenly disconnected from the power source. What they do following that is called the Natural response, which is defined by First Order Differential Equations.

“n” here means for any quantity of resistors, capacitors or inductors that are all in either series of parallel.

Component:ResistanceCapacitanceInductance
Series:R1+R2+Rn=R1/C1+1/C2+1/Cn=1/CL1+L2+Ln=L
Parallel:1/R1+1/R2+1/Rn=1/RC1+C2+Cn=C1/L1+1/L2+1/Cn=1/L
EquationR=V/IC=i(t)/(dv/dt)L=v(t)/(di/dt)

Types of circuits RLC, RL, RC circuits

Natural response vs step response

There are various current equations for a Resistor, Inductor and Capacitor.

IR=V/RI_R=V/R
IL=1/L∗∫0tvdtI_L=1/L*\int_0^t{v}dt
IC=CdvdtI_C=C\frac{dv}{dt}

There are various voltage equations for a Resistor, Inductor and Capacitor.

VR=IRV_R=IR
VL=LdidtV_L=L\frac{di}{dt}
VC=1C∫0tidtV_C=\frac1C\int_0^t{idt}

RL First Order Differential Equation

RC First Order Differential Equation


RL Power Equation

RC Power Equation

Time Constant

RC Example:

RL Example

From these we can get two differential equations which are straight forward to solve.

RC

RL

Section 9: Thevenin and Norton Equivalent Circuits.

Thevenin and Norton Equivalent circuits are ways to compress a more complicated circuit into a single equivalent Resistor and Voltage/Current Source. If you need to test various different components this can greatly simplify the system.

Thevenin Equivalent Steps

  1. Open the Load Resistor (Remove it from the circuit)
  2. Find the equivalent voltage for the given components Terminals
  3. Short all the Voltage Sources and Open all the Current Sources.
  4. Find the equivalent Resistance.

Example:

Norton Equivalent Steps

  1. Open the Load Resistor (Remove it from the circuit)
  2. Find the equivalent current for the given wire left by the Short.
  3. Short all the Voltage Sources and Open all the Current Sources.
  4. Find the equivalent Resistance.

Example:

Section #: AC Current

Sinusoidal Sources

VRMS

Steady Stage Response

Section #: Phasors

Capacitors and Inductors

Super Nodes and

Voltage and Current Source conversion

Thevenin’s Theorem

Thevenin Equivalent

OpAmps

Section 2: Capacitors, Inductors and Step Response

Section 3: AC Circuits and Phasors

Filter Circuits

Capacitors and Inductors

Response

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