Electromotive Force, Ohm's Law, and Electrical Circuit Analysis Study Guide

Electromotive Force and Potential Difference

Units Used For Calculations

A device that has the ability to maintain a potential difference in charge between two points is said to develop an electromotive force (EMF).

Potential Difference (p.d.) is the difference in electric charge between two points and is measured in Volt ($V$).

Key Concepts of Voltage:

• A potential difference causes a current to flow.

• An EMF maintains a potential difference.

• Although potential difference, EMF, and Voltage do NOT mean exactly the same thing conceptually, in calculations, the symbols $E$ or $V$ are used interchangeably for voltage, EMF, and potential difference.

Direct Current Circuits and Ohm's Law

Simple Direct Current Circuits

At its simplest, an electric circuit consists of:

1. A source of electromotive force.

2. A wire or conductor connecting the source to a load or resistance ($R$).

3. A second wire connecting the load back to the source again.

In circuit diagrams, the following notation is used:

• $I$: Current in Ampere ($A$).

• $E$: Electromotive force in Volt ($V$).

• $R$: Resistance in Ohm ($\Omega$).

Ohm's Law

Ohm's Law states that current ($I$) is directly proportional to electromotive force ($E$) and inversely proportional to resistance ($R$).

Mathematically, it can be written as: $I = \frac{E}{R}$

Where:

• $I$ = Current in Ampere ($A$)

• $E$ = Electromotive Force (Potential Difference) in Volt ($V$)

• $R$ = Resistance in Ohm ($\Omega$)

The equation can be algebraically rearranged as follows:

• To find EMF: $E = I \times R$

• To find Resistance: $R = \frac{E}{I}$

Kirchhoff's Laws

First Law – Kirchhoff's Current Law (KCL)

• The algebraic sums of the currents at any electrical junction (node) must equal zero.

• Alternatively stated: The sum of currents flowing away from any point in an electric circuit must equal the sum of currents flowing toward that point.

Example Scenario (Fig. 2): At a junction where currents $I_1$ and $I_4$ flow into the node and $I_2$, $I_3$, and $I_5$ flow away from it, the equation is: $I_1 + I_4 = I_2 + I_3 + I_5$

Second Law – Kirchhoff's Voltage Law (KVL)

Around any closed path in an electric circuit, the algebraic sum of all potential differences (voltages) is zero.

Steps to Apply Kirchhoff's Second Law:

1. Specify the direction of the different EMF and voltage drops.

2. Note: We will consider the direction of potential difference (change of voltage) positive if it is in the clockwise direction.

3. The direction of the voltage drop across a resistance is the same as the direction of current flow through the resistance.

4. Mark the direction of current in every branch (connection between two junctions).

5. Voltage drop across a resistor is calculated as: $E_R = I \times R$

Kirchhoff's Laws Analysis Examples

Example: Calculating Current Junctions and Closed Paths

Given a circuit with junctions A and B, and branches containing resistors $R_1$, $R_2$, $R_3$ and EMF sources $E_1$, $E_2$:

Current Law Equations:

• For junction A: $I_1 = I_2 + I_3$

• For junction B: $I_2 + I_3 = I_1$

Voltage Law Equations (Paths):

• Path 1:   $V_{R1} + V_{R2} + (-E_1) = 0$   $(I_1 \times R_1) + (I_2 \times R_2) + (-E_1) = 0$

• Path 2:   $(-E_2) + V_{R3} + (-V_{R2}) = 0$   $(-E_2) + (I_3 \times R_3) + (-I_2 \times R_2) = 0$

• Path 3:   $V_{R1} + (-E_2) + V_{R3} + (-E_1) = 0$   $(I_1 \times R_1) + (-E_2) + (I_3 \times R_3) + (-E_1) = 0$

Series Circuits

Characteristics of Resistors in Series:

1. The same current flows through all resistors in the series.

2. The total resistance ($R_T$) is the sum of all individual resistances:    $R_T = R_1 + R_2 + R_3 + ...$

3. The sum of the voltage drops across individual resistors equals the total potential (voltage) drop in the circuit ($V_T$).

4. The supplied EMF ($E$) is equal to the total potential drop ($V_T$).

Calculation Summary for Circuit Analysis:

• Individual Voltage Drop ($V_1$): $V_1 = I \times R_1$

• Total Voltage ($V_T$): $V_T = V_1 + V_2 + V_3$

• Substitution Method: $V_T = I \times (R_1 + R_2 + R_3)$

Note: The resistance of the wires from the generator and the resistors is typically assumed to be zero unless explicitly included as a specific resistor (e.g., $R_4$).

Series Circuit Examples

Example 1: Three Resistors Given: $R_1 = 4\,\Omega$, $R_2 = 6\,\Omega$, $R_3 = 8\,\Omega$, and $I = 2\,A$.

(a) Total Resistance ($R_T$): $R_T = 4\,\Omega + 6\,\Omega + 8\,\Omega = 18\,\Omega$

(b) Voltage Drop Across Each:

• $V_1 = 2\,A \times 4\,\Omega = 8\,V$

• $V_2 = 2\,A \times 6\,\Omega = 12\,V$

• $V_3 = 2\,A \times 8\,\Omega = 16\,V$

(c) Total Voltage Drop: $V_T = 8\,V + 12\,V + 16\,V = 36\,V$ OR $V_T = I \times R_T = 2\,A \times 18\,\Omega = 36\,V$

Example 2: Four Resistors Given: $R_1 = 5\,\Omega$, $R_2 = 4\,\Omega$, $R_3 = 10\,\Omega$, $R_4 = 6\,\Omega$, and $E = 50\,V$.

(a) Total Resistance ($R_T$): $R_T = 5\,\Omega + 4\,\Omega + 10\,\Omega + 6\,\Omega = 25\,\Omega$

(b) Current ($I$): $I = \frac{E}{R_T} = \frac{50\,V}{25\,\Omega} = 2\,A$

(c) Voltage Drops:

• $V_1 = 2\,A \times 5\,\Omega = 10\,V$

• $V_2 = 2\,A \times 4\,\Omega = 8\,V$

• $V_3 = 2\,A \times 10\,\Omega = 20\,V$

• $V_4 = 2\,A \times 6\,\Omega = 12\,V$

Check: $10\,V + 8\,V + 20\,V + 12\,V = 50\,V$ (Matches supplied EMF).

Example 3: Unknown Resistance Given: $I = 5\,A$, $E = 110\,V$, $R_1 = 3\,\Omega$, $R_2 = 5\,\Omega$, $R_3 = 2\,\Omega$, $R_4 = 8\,\Omega$, and $R_5$ is unknown.

(a) Total Resistance ($R_T$): $R_T = \frac{E}{I} = \frac{110\,V}{5\,A} = 22\,\Omega$

(b) Solve for $R_5$: $R_5 = R_T - R_1 - R_2 - R_3 - R_4$ $R_5 = 22 - 3 - 5 - 2 - 8 = 4\,\Omega$

(c) Voltage Drops:

• $V_1 = 5\,A \times 3\,\Omega = 15\,V$

• $V_2 = 5\,A \times 5\,\Omega = 25\,V$

• $V_3 = 5\,A \times 2\,\Omega = 10\,V$

• $V_4 = 5\,A \times 8\,\Omega = 40\,V$

• $V_5 = 5\,A \times 4\,\Omega = 20\,V$

Verification: $15 + 25 + 10 + 40 + 20 = 110\,V$.

Parallel Circuits

Characteristics of Parallel Circuits:

1. The sum of the individual current flows through each loop/branch is equal to the total current in the circuit.

   • Contrast: This differs from series circuits where the same current flows through all components.

2. In a parallel circuit, the same voltage is applied to all resistors.

Equivalent Resistance Formula: Using Ohm's Law where $I_T = I_1 + I_2 + I_3 + ...$ ${I_T} = \frac{E}{R_1} + \frac{E}{R_2} + \frac{E}{R_3}$ $I_T = E \times [\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}]$

Dividing both sides by the total voltage yields the total resistance formula: $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$

Parallel Circuit Example

Given components: $R_1 = 5\,\Omega$, $R_2 = 8\,\Omega$, $R_3 = 6\,\Omega$, $R_4 = 16\,\Omega$. Voltage = $100\,V$.

(a) Equivalent Resistance ($R_T$): $\frac{1}{R_T} = \frac{1}{5} + \frac{1}{8} + \frac{1}{6} + \frac{1}{16}$ $\frac{1}{R_T} = 0.2 + 0.125 + 0.167 + 0.0625 = 0.5545$ $R_T = \frac{1}{0.5545} = 1.805\,\Omega$

(b) Total Current ($I_T$): $I_T = \frac{100\,V}{1.805\,\Omega} = 55.40\,A$

(c) Branch Currents:

• $I_1 = \frac{100}{5} = 20\,A$

• $I_2 = \frac{100}{8} = 12.5\,A$

• $I_3 = \frac{100}{6} = 16.67\,A$

• $I_4 = \frac{100}{16} = 6.25\,A$

Check: $20 + 12.5 + 16.67 + 6.25 = 55.42\,A$

Wheatstone Bridge

Theoretical Foundation Consider a supply $E = 120\,V$ feeding two parallel branches. Branch 1 (ABC): $R_1 = 10\,\Omega$, $R_2 = 14\,\Omega$.

• Current ($I_1$) = $\frac{120}{10 + 14} = 5\,A$.

• Voltage drop AB = $5\,A \times 10\,\Omega = 50\,V$.

• Voltage drop BC = $5\,A \times 14\,\Omega = 70\,V$.

Branch 2 (ADC): $R_3 = 24\,\Omega$, $R_4 = 16\,\Omega$.

• Current ($I_2$) = $\frac{120}{16 + 24} = 3\,A$.

• Voltage drop AD = $3\,A \times 16\,\Omega = 48\,V$.

A galvanometer (acting as a voltmeter) placed across points B and D would show a potential difference of $2\,V$ ($50\,V - 48\,V$).

Balanced Condition A bridge is balanced when the potential at points B and D are equal, causing the galvanometer to indicate zero.

Mathematical derivation for balance: $I_1 \times R_1 = I_2 \times R_4$ $I_1 \times R_2 = I_2 \times R_3$

Dividing these equations: $\frac{R_1}{R_2} = \frac{R_4}{R_3}$

General product rule for balance: $R_1 \times R_3 = R_2 \times R_4$

Applications:

• Measuring unknown resistance ($R_x$) by adjusting a variable known resistance until the bridge is balanced.

• Ohmmeters frequently utilize Wheatstone Bridge arrangements.

• Measurement of temperature using a thermistor (temperature-sensitive resistor) in one arm of the bridge.

Energy Principles

Defining Energy Energy is the ability to do work. It is essentially stored work. Examples:

• A coiled spring powering a clock.

• Water stored behind a dam (potential energy released to generate mechanical and then electrical energy via turbines/generators).

Conservation of Energy

• Energy can neither be created nor destroyed; this is the Principle of Conservation of Energy.

• Transformation example: An electric motor changes electrical energy into mechanical energy and heat energy.

Electrical Work and Energy

Work and Units

• Work is energy transferred when a force moves through a distance.

• One Joule ($J$) is the work done when a force of one Newton moves through a distance of one meter ($1\,Nm$).

• Power is the rate of energy transfer.

• Unit of power is the Watt ($W$). $1\,Watt = 1\,\frac{Joul}{second}$.

Work Calculation Formulas Work done depends on current flow, voltage, and the duration of time the current flows through the load.

Basic formula: $W = E \times I \times t$

Where:

• $W$ = Work in Watt seconds (Joules)

• $E$ = Potential difference in Volt

• $I$ = Current in Ampere

• $t$ = Time in seconds ($s$)

Alternative formulas (combining Ohm's Law and Work):

• $W = I^2 \times R \times t$

• $W = \frac{E^2}{R} \times t$

Energy Units of Measure

• Watt second ($W \cdot s$) or Joule ($J$) is a small unit.

• Watt-hour ($Wh$): $1\,Wh = 3,600\,Ws = 3,600\,J$

• Kilowatt-hour ($kWh$): $1\,kWh = 10^3\,Wh = 3.6 \times 10^6\,J = 3.6\,MJ$

Electrical Power

Power ($P$) is the rate of doing work or transferring energy. $Power (P) = \frac{Work (W)}{Time (t)}$

Formulas for Power:

• $P = E \times I$

• $P = I^2 \times R$

• $P = \frac{E^2}{R}$

Power Rating of Resistors

• The power rating is the amount of power a resistor can dissipate without affecting its characteristics.

• Power absorbed by a resistance is changed to heat.

• Ratings are typically related to a specific temperature, such as $20\,^\circ C$.

Electrical Power Examples

Example 1: Electric Lamp Lamp draws $1\,A$ from a $120\,V$ line. $P = 120\,V \times 1\,A = 120\,W$

Example 2: Electric Heater Heater with $30\,\Omega$ resistance and $240\,V$ application. $P = \frac{(240\,V)^2}{30\,\Omega} = \frac{57,600}{30} = 1,920\,W = 1.92\,kW$

Example 3: Finding Resistance from Energy Usage Heater used $10\,kWh$ in $8\,h$ at $120\,V$.

1. Find Power ($P$): $P = \frac{10\,kWh}{8\,h} = 1.25\,kW$

2. Find Resistance ($R$): $R = \frac{E^2}{P} = \frac{(120\,V)^2}{1,250\,W} = \frac{14,400}{1,250} = 11.52\,\Omega$

Example 4: Calculating Joules A $100\,W$ lamp used for $12\,h$.

• Energy in $Wh$: $100\,W \times 12\,h = 1,200\,Wh = 1.2\,kWh$

• Energy in $MJ$: $1.2\,kWh \times 3.6\,MJ/kWh = 4.32\,MJ$

• Energy in Joules: $4.32 \times 10^6\,J$

Alternating Current (AC) Principles

Most electrical power is supplied as alternating current due to two advantages:

1. It can be generated without commutator limits.

2. Voltage can be easily transformed up or down for transmission and distribution.

Generation and Sine Waves

• A conductor rotated through a magnetic field produces an alternating EMF shaped like a sine wave.

• One rotation across two poles produces one cycle (Zero $\rightarrow$ Positive Maximum $\rightarrow$ Zero $\rightarrow$ Negative Maximum $\rightarrow$ Zero).

• The number of cycles per second determines the Frequency, measured in Hertz ($Hz$).

Phase Relationships in AC Circuits

• In Phase: If voltage and current reach maximum values at the same time. This occurs in purely resistive circuits.

• Lagging Current: Current reaches maximum later than voltage.

• Leading Current: Current reaches maximum earlier than voltage.

• Out of Phase: Maximum and zero values do not occur at the same time.

AC Generation, Cycle, and Frequency

• Instantaneous Value: The value of EMF at any specific given time ($e_1$, $i_1$).

• Period ($T$): The time required for one full cycle, measured in seconds ($s$).

• Frequency ($f$): Cycles per second ($Hz$).

• Common Power Frequencies: $60\,Hz$ in America; $50\,Hz$ in Europe, most of Asia, and Africa.

Relationship: $f = \frac{1}{T}$

Example: Wave with period of $2\,\mu s$ ($2 \times 10^{-6}\,s$): $f = \frac{1}{2 \times 10^{-6}\,s} = 0.5 \times 10^6\,Hz = 0.5\,MHz$

Practice Problems:

• Wave with period $4\,\mu s$: Frequency = $0.25\,MHz$

• $50\,Hz$ wave: Period = $0.02\,s$

Values of Alternating Current and Voltage

AC voltages and currents have multiple types of values:

1. Maximum Value ($I_{max}$ or $E_{max}$): Also called Amplitude. The largest value in a positive or negative direction.

2. Peak-to-Peak Value ($I_{pp}$ or $E_{pp}$): Magnitude between the highest positive and lowest negative peaks. It is twice the maximum value.    $I_{pp} = 2 \times I_{max}$

3. Instantaneous Value: Changing values throughout the cycle.

4. Average Value: (Not described in detail in text but noted as a type).

5. Effective Value (RMS): Based on the square of the current flow producing heat. It is the Root Mean Square of the instantaneous values.

Root Mean Square (RMS) Values

• The effective (RMS) value of a pure sine wave is 0.707 times the peak value.

• RMS is used because the heating effect ($I^2 \times R$) is dependent on the square of the current.

• Current and voltage in AC are nearly always quoted as RMS values.

Formulas:

• $E_{RMS} = 0.707 \times E_{max}$

• $I_{RMS} = 0.707 \times I_{max}$

Example 1: Maximum value = $170\,V$.

• $E_{pp} = 2 \times 170\,V = 340\,V$

• $E_{RMS} = 0.707 \times 170\,V = 120.19\,V$

Example 2: RMS value = $12\,A$.

• Find Max: $I_{max} = \frac{12\,A}{0.707} = 16.9731\,A$

• Find Peak-to-Peak: $I_{pp} = 2 \times 16.9731\,A = 33.94625\,A$

Inductance and Inductive Reactance

Inductance ($L$)

• A biological circuit property involving coils, often wrapped around magnetic cores.

• Opposition to change in current flow.

• Only affects current when the value is changing (always changing in AC).

• Unit: Henry ($H$).

Inductive Reactance ($X_L$)

• The specific opposition of inductance to AC current flow, measured in Ohm ($\Omega$).

• Formulas:   $I = \frac{E}{X_L}$   $X_L = 2 \times \pi \times f \times L$

• Relationship: Current lags applied voltage by $90^\circ$ in a purely inductive circuit.

Example: Coil $L = 0.2\,H$, connected to $120\,V$, $60\,Hz$ supply.

• $X_L = 2 \times 3.1416 \times 60 \times 0.2 = 75.4\,\Omega$

• $I = \frac{120\,V}{75.4\,\Omega} = 1.59\,A$

Capacitance and Capacitive Reactance

Capacitance ($C$)

• Formed by two conductors separated by an insulator (dielectric).

• Current flows only if applied voltage is changing.

• Units: Farad ($F$). Common units: $\mu F$ ($10^{-6}$) and $pF$ ($10^{-12}$).

Capacitive Reactance ($X_c$)

• The opposition of the capacitor to AC flow.

• Formulas:   $I = \frac{E}{X_c}$   $X_c = \frac{1}{2 \times \pi \times f \times C}$

• Relationship: Current leads applied voltage by $90^\circ$ in a purely capacitive circuit.

Example: $0.2\,\mu F$ capacitor at $60\,Hz$ and $600\,kHz$.

• At $60\,Hz$: $X_c = \frac{1}{2 \times \pi \times 60 \times 0.2 \times 10^{-6}} = 13.3\,k\Omega$

• At $600\,kHz$: $X_c = \frac{1}{2 \times \pi \times 6 \times 10^5 \times 0.2 \times 10^{-6}} = 1.33\,\Omega$

Combined Reactance and Impedance

Combined Reactance ($X$) Since inductive and capacitive reactances have opposite effects (leading vs lagging phase), their effects tend to cancel. The combined effect is the difference: $X = X_L - X_c$

Impedance ($Z$) Impedance is the total opposition to current flow consisting of resistance and reactance. Unit: Ohm ($\Omega$).

Formula for a series circuit: $Z = \sqrt{R^2 + (X_L - X_c)^2}$

Example: $R = 50\,\Omega$, $X_L = 70\,\Omega$, $X_c = 20\,\Omega$. $Z = \sqrt{50^2 + (70 - 20)^2}$ $Z = \sqrt{2500 + 2500} = \sqrt{5000} = 70.71\,\Omega$

Power in AC Systems

Types of Power:

1. Apparent Power ($S$): Product of effective current and voltage. Expressed in VoltAmperes ($VA$) or $kVA$.    $S = E \times I$

2. Active Power / Real Power ($P$): True power used in resistive components. Expressed in Watt ($W$) or $kW$.    $P = E \times I \times \cos(\theta)$

3. Reactive Power / Imaginary Power ($A$ or $Q$): Power utilized by inductors and capacitors. Energy is taken for part of the cycle and returned in another. Net power consumed is Zero. Measured in VAR (VoltAmpere Reactive).    $A = E \times I \times \sin(\theta)$

Phase Angle ($\theta$): The angle by which current leads or lags voltage.

Power Factor (pf)

Definition

• Power Factor is the ratio of real power to apparent power.

• It is represented by $\cos(\theta)$, where $\theta$ is the phase angle between EMF and current.

• Value ranges between 0.0 and 1.0 (or 0% to 100%).

Formula Relationship: $Active\,Power (W) = S \times pf$ $VA = \sqrt{(Active\,Power)^2 + (Reactive\,Power)^2}$

Power Factor Improvement: Low power factor (e.g., 75% due to induction motors) results in higher line losses and voltage drops. Corrective capacitors are used to improve (increase) the power factor.

Power Factor Calculation Examples

Example 1: Solving for Real and Imaginary Power Given: $E = 220\,V$, $I = 20\,A$, $pf = 78.8\%$ ($0.788$).

(a) Real Power ($P$): $P = 220\,V \times 20\,A \times 0.788 = 3,467\,W \approx 3.467\,kW$

(b) Imaginary Power ($A$):

1. Find $\theta$: $\cos(\theta) = 0.788 \implies \theta = 38^\circ$

2. Find $\sin(\theta)$: $\sin(38) = 0.6157$

3. Calculate $A$: $A = 220 \times 20 \times 0.6157 = 2,709\,VAR \approx 2.709\,kVAR$

Example 2: Determining pf from Meter Readings Given: Wattmeter = $2,400\,W$, Voltmeter = $240\,V$, Ammeter = $15\,A$, Freq = $60\,Hz$.

(a) Apparent Power ($S$): $S = 240\,V \times 15\,A = 3,600\,VA = 3.6\,kVA$

(b) Power Factor: $pf = \frac{P}{ExI} = \frac{2,400\,W}{3,600\,VA} = 0.67$ or $67\%$

Three-Phase Circuits

General Principles

• Balanced three-phase circuits combine three single-phase circuits.

• Three phase voltages and currents are spaced by 120 degrees.

• Results in a smoother, less pulsating flow of power compared to single-phase.

Three-Phase Connections

1. Star (Wye) Connection:

   • One end of each of the three windings joins at a central "Star Point" (Neutral).

   • Line Voltage ($E_L$) = $\sqrt{3} \times Phase\,Voltage (E_p)$.

   • Line Current ($I_L$) = Phase Current ($I_p$).

   • Example: Phase voltage $100\,V$ implies line voltage $100 \times 1.732 = 173\,V$.

2. Delta Connection:

   • Windings connected in a closed loop; no neutral point available.

   • Line Voltage ($E_L$) = Phase Voltage ($E_p$).

   • Line Current ($I_L$) = $\sqrt{3} \times Phase\,Current (I_p)$.

   • Example: Phase current $1\,A$ implies line current $1 \times 1.732 = 1.732\,A$.

Power in Three-Phase Circuits

Regardless of Star or Delta connection, the formula for power using line-to-line values is identical:

Real Power ($P$): $P = \sqrt{3} \times E_L \times I_L \times \cos(\theta)$

Apparent Power ($S$): $S = \sqrt{3} \times E_L \times I_L$

Example Problem: Generator: $E_L = 480\,V$, full load current $I_L = 300\,A$, lagging power factor $75\%$.

(a) Apparent Power ($S$): $S = 1.732 \times 480\,V \times 300\,A = 249.12 \times 10^3\,VA = 249.12\,kVA$

(b) Real Power ($P$): $P = 249.12\,kVA \times 0.75 = 186.84\,kW$

(c) Reactive Power ($A$):

1. $\cos(\theta) = 0.75 \implies \theta = 41.409^\circ$

2. $\sin(41.409) = 0.6614$

3. $A = S \times \sin(\theta) = 249.12 \times 0.6614 = 164.78\,kVAR$