Physics 11 Unit 7: Electric Circuits Comprehensive Study Notes

Physics 11: Unit 7 - Electric Circuits

Learning Outcomes

By the end of this study guide, you should be able to:

• Draw and define the schematics of an electric circuit.

• Identify the difference between a series and parallel circuit.

• Use Ohm's Law to calculate the Voltage, Current, and Resistance of a circuit.

• Use Kirchhoff's Law to determine Voltage, Current, and Resistance of a circuit.

• Determine the Electromotive Force (EMF) of a circuit.

• Determine the cost of electric power.

Lesson 1: Calculating Voltage, Current, and Resistance

Current Electricity Basics

• Definition: Current electricity is defined as the flow of electrons through a conductor.

• Current ($I$): The specific quantity describing the number of charges flowing per second.

• Units: Amperes or "Amps" ($A$).

• Formula: $I = \frac{q}{t}$

• Where:

  • $I$ = Current ($A$)

  • $q$ = Charge ($C$ - Coulombs)

  • $t$ = Time ($s$)

Charge and Electrons

• To find the number of electrons, use the elementary charge constant: $1e^- = 1.6 \times 10^{-19}\,C$

Rules for Current Flow

For current to flow through a conductor, two conditions must be met:

1. A Potential Difference: Provided by a voltage source.

2. A Complete Circuit: A continuous path for the electrons to travel.

Voltage Sources

Examples of everyday voltage sources include:

• Batteries (Cells)

• Electrical outlets

Analogies and Conceptual Notes

• The River Analogy: Consider a river. The rate of water flowing down the river represents the current. Note that current refers to the rate of water flowing, not the individual speed of water molecules. In electric circuits, current represents how many electrons pass a certain point in a specific amount of time.

Voltage ($V$)

• The units of voltage are Volts ($V$).

Resistance ($R$)

• Resistance is the opposition to the flow of current. The units of resistance are Ohms ($\Omega$).

Ohm's Law

The three quantities of Current, Voltage, and Resistance are related by the following formula: $V = IR$

Conventional Current vs. Electron Flow

There are two standards used to describe the direction of current flow in a circuit:

1. Electron Flow: This represents the actual physical movement of electrons. Electrons flow from the negative terminal through the circuit and into the positive terminal.

2. Conventional Current: This is defined as the flow of positive charge. Positive charges flow from the positive terminal to the negative terminal.

Usage Standards

• History: Conventional current was established during the discovery of electricity before electrons were understood. It was later found to be physically incorrect regarding the particles moving, but the convention remains.

• High School/Technical Programs: Generally use electron flow.

• University Courses: Conventional current is the preferred method.

• Class Rule: Unless otherwise stated, this course uses Conventional Current ($+\text{ to } -$). It is vital to remain consistent with the chosen method to avoid confusion.

Power ($P$)

Power is often confused with voltage or energy, but it has a specific definition in physics.

Definitions and Equations

• General Definition: Power is the rate of doing work.

• General Formula: $P = \frac{W}{t} = \frac{\Delta E}{t}$

• Electric Power Formula: $P = IV$

Alternative Power Formulas

By substituting Ohm's Law ($V = IR$) into the electric power equation, we can derive:

1. $P = I^2 R$

2. $P = \frac{V^2}{R}$

Example Problems: Lesson 1

1. Electric Fan Calculation:

   • Given: $R = 12\,\Omega$, $I = 0.75\,A$.

   • Find: Voltage ($V$).

   • Solution: $V = IR = (0.75\,A)(12\,\Omega) = 9\,V$.

2. Electric Heater Resistance:

   • Given: $P = 1.00 \times 10^2\,W$, $V = 120\,V$.

   • Find: Resistance ($R$).

   • Solution: $P = \frac{V^2}{R} \rightarrow R = \frac{V^2}{P} = \frac{(120\,V)^2}{100\,W} = 144\,\Omega$.

3. Blender Calculations:

   • Given: $I = 0.25\,A$, $V = 120\,V$, $t = 15\,s$.

   • a) Resistance: $R = \frac{V}{I} = \frac{120\,V}{0.25\,A} = 480\,\Omega$.

   • b) Power: $P = IV = (0.25\,A)(120\,V) = 30\,W$.

   • c) Electron Count:

     • Find charge: $q = It = (0.25\,A)(15\,s) = 3.75\,C$.

     • Convert to electrons: $\frac{3.75\,C}{1.6 \times 10^{-19}\,C/e^-} = 2.34 \times 10^{19} \text{ electrons}$.

Lesson 2: Schematics of an Electric Circuit

Schematic Symbols and Functions

• Wire: Used for the transfer of current.

• Open Switch: Stops the flow of current in a wire.

• Closed Switch: Allows the flow of current in a wire.

• Resistor: Resists the flow of current.

• Single Cell: Source of voltage with potential difference.

• 3-Cell Battery: A combined source of voltage with potential difference.

• Ammeter ($A$): Measures current in a circuit.

• Voltmeter ($V$): Measures voltage at a point in a circuit.

Circuit Configurations

1. Series: There is only one path for current to flow. Components are connected end-to-end.

2. Parallel: There are multiple paths for current to flow. Components are connected across common junctions.

Measuring Voltage and Current

• Voltmeter: Must be connected in Parallel. This is because it measures the voltage drop across a device.

• Ammeter: Must be connected in Series. This is because it measures the current through a circuit.

Types of Current

• DC (Direct Current): Current that flows in only one direction, such as that from a battery.

• AC (Alternating Current): Current that alternates its direction of flow. This is the power supplied to homes. In North America, the frequency is $60\,Hz$. Physics 11 focuses exclusively on DC.

Lesson 3: Basic Circuits (Series and Parallel Rules)

Series Circuits

• Path: One path for electrons.

• Current: $I_T = I_1 = I_2 = I_3$.

• Voltage: $V_T = V_1 + V_2 + V_3$.

• Resistance: $R_T = R_1 + R_2 + R_3$. (Total resistance increases as resistors are added).

Parallel Circuits

• Path: More than one path for electrons.

• Current: $I_T = I_1 + I_2 + I_3$.

• Voltage: $V_T = V_1 = V_2 = V_3$.

• Resistance: $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$. (Total resistance decreases as resistors are added).

Solving Tips for Circuit Problems

1. Draw the circuit diagram if it is not provided.

2. Label each resistor with its $V, I, R$ and the battery with its $V_T, I_T, R_T$.

3. Apply Series and Parallel rules. When two of the three variables ($V, I, R$) are known for a component, use Ohm's Law ($V=IR$) to find the third.

4. Equivalent Resistance ($R_{eq}$): For complex circuits, transform combination branches into a single equivalent resistance to simplify the circuit into a basic series circuit.

Example Problems: Lesson 3

1. Equivalent Resistance (Mixed Circuit):

   • Circuit: A $20\,\Omega$ resistor in series with a parallel branch containing a $10\,\Omega$ and a $5\,\Omega$ resistor.

   • Step 1: Find $R_{eq}$ for parallel part: $\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{5} = \frac{1}{10} + \frac{2}{10} = \frac{3}{10} \rightarrow R_{eq} = 3.33\,\Omega$.

   • Step 2: Add series resistance: $R_T = 20 + 3.33 = 23.33\,\Omega$.

2. Equivalent Resistance (Parallel Series Branches):

   • Circuit: Two parallel branches. Branch A has $1.5\,k\Omega + 4.5\,k\Omega$ in series. Branch B has $2\,k\Omega + 4\,k\Omega$ in series.

   • Step 1: Branch A $R = 6\,k\Omega$; Branch B $R = 6\,k\Omega$.

   • Step 2: Total resistance: $\frac{1}{R_T} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \rightarrow R_T = 3\,k\Omega$ (or $3000\,\Omega$).

3. Current in Parallel:

   • Given: $10\,V$ parallel circuit with a $3\,\Omega$ and a $5\,\Omega$ resistor. Find current through the $5\,\Omega$ resistor.

   • Solution: Since it is parallel, $V_2 = V_T = 10\,V$. Therefore, $I_2 = \frac{V_2}{R_2} = \frac{10\,V}{5\,\Omega} = 2\,A$.

4. Finding unknown Resistance in Series:

   • Given: $V_T = 12\,V, I_T = 0.3\,A, R_1 = 12\,\Omega, R_2 = 18\,\Omega$. Find $R_3$.

   • Step 1: Find $R_T = \frac{V_T}{I_T} = \frac{12\,V}{0.3\,A} = 40\,\Omega$.

   • Step 2: $R_3 = R_T - R_1 - R_2 = 40 - 12 - 18 = 10\,\Omega$.

5. Complex Circuit Solution:

   • Circuit: $20\,V$ battery. $R_3 = 8\,\Omega$ is in series with a parallel branch of $R_1 = 6\,\Omega$ and $R_2 = 18\,\Omega$.

   • Step 1: $R_{eq} \text{ (parallel branch)} = (\frac{1}{6} + \frac{1}{18})^{-1} = (\frac{4}{18})^{-1} = 4.5\,\Omega$.

   • Step 2: $R_T = 8 + 4.5 = 12.5\,\Omega$.

   • Step 3: $I_T = \frac{20\,V}{12.5\,\Omega} = 1.6\,A$. Therefore, $I_3 = 1.6\,A$.

   • Step 4: $V_{eq} = I_T \times R_{eq} = (1.6\,A)(4.5\,\Omega) = 7.2\,V$. Because it's parallel, $V_1 = V_2 = 7.2\,V$.

   • Step 5: $I_1 = \frac{7.2\,V}{6\,\Omega} = 1.2\,A$. $I_2 = \frac{7.2\,V}{18\,\Omega} = 0.4\,A$.

Lesson 4: Kirchhoff's Laws

Kirchhoff's Current Law (Junction Rule)

• Definition: The sum of currents entering a junction must equal the sum of currents leaving a junction.

• Junction: A point where two or more things join or split.

• In Series: Current is the same everywhere ($I_T = I_1 = I_2 = I_3$).

• In Parallel: Current splits. The current in each path adds to the total ($I_T = I_1 + I_2 + I_3$).

Kirchhoff's Voltage Law (Loop Rule)

• Definition: For any closed loop, the sum of voltage gains is equal to the sum of voltage drops. This is a restatement of the Law of Conservation of Energy.

• Gains: Occur across the terminals of a cell/battery.

• Drops: Occur across resistors.

• In Series: $V_T = V_1 + V_2 + V_3$.

• In Parallel: Potential difference is the same across each resistor ($V_T = V_1 = V_2 = V_3$).

Resistance Analogies

• Series: Each electron must push through each resistor, increasing total resistance.

• Parallel: Like adding cash registers in a store. More cashiers (resistors in parallel) reduce the overall resistance to the flow of customers (current).

Lesson 5: Electromotive Force (EMF) and Terminal Voltage

Definitions

• EMF ($\varepsilon$): The potential difference between terminals when the battery is not connected to a circuit. Despite the name "Electro Motive Force," it is a Voltage, not a force.

• Internal Resistance ($r$): Every battery has some resistance inside it. When current flows, some voltage is dropped internally.

• Terminal Voltage ($V_t$): The actual voltage available to the external circuit when current is flowing. It is always less than EMF ($\varepsilon$).

Equations

• Discharging (Standard Use): $V_t = \varepsilon - Ir$

• If not connected ($I = 0$): $V_t = \varepsilon$.

• Charging: To force electrons backwards into a battery, the external voltage must be larger. $V_t = \varepsilon + Ir$

Example Problems: Lesson 5

1. Terminal Voltage Calculation:

   • Given: $\varepsilon = 12.0\,V, r = 0.220\,\Omega, I = 3.00\,A$.

   • Solution: $V_t = 12.0 - (3.00)(0.220) = 11.34\,V$.

2. Charging a Battery:

   • Given: $\varepsilon = 12.0\,V, V_{external} (V_t) = 15\,V, r = 1.3\,\Omega$.

   • Find: Current ($I$).

   • Solution: $V_t = \varepsilon + Ir \rightarrow I = \frac{V_t - \varepsilon}{r} = \frac{15 - 12}{1.3} = 2.3\,A$.

3. Circuit with Internal Resistance:

   • Given: $\varepsilon = 6.00\,V, r = 0.50\,\Omega, R_{load} = 4.00\,\Omega$.

   • Step 1: $R_T = R + r = 4.00 + 0.50 = 4.50\,\Omega$.

   • Step 2: $I_T = \frac{\varepsilon}{R_T} = \frac{6.00\,V}{4.50\,\Omega} = 1.33\,A$.

   • Step 3: $V_t = \varepsilon - Ir = 6.00 - (1.33)(0.50) = 5.34\,V$.

Lesson 6: Determining Cost of Power

Energy Conversion and Power

• Devices convert electrical energy into heat, light, sound, or motion.

• Power ($P$): The rate at which energy is converted. Units are Watts ($W$).

• $1\,Watt = 1\,Joule/second$.

Kilowatt-Hours ($kWh$)

• Utilities charge based on energy consumption ($E$) in kilowatt-hours ($kWh$).

• Definition: A $kWh$ is the amount of energy used by a $1000\,W$ ($1\,kW$) device running for $1\,hour$.

Cost Calculation

• Factors: Power rating, duration of use, utility rate.

• Formula: $\text{Cost} = \text{Energy Used (kWh)} \times \text{Rate}$

BC Hydro Tiered System (Example Data)

• Step 1 Rate: $11.87\,\text{cents}/kWh$ (applies to the first $1,376\,kWh$ in a 2-month period).

• Step 2 Rate: $14.08\,\text{cents}/kWh$ (applies to usage beyond Step 1).

Example Problems: Lesson 6

1. Clothes Dryer:

   • Given: $240\,V, 18\,A, 2.5\,hours$. Rate: $11.87\,\text{cents}/kWh$.

   • Power: $P = IV = (240)(18) = 4320\,W = 4.32\,kW$.

   • Energy: $E = P \times t = (4.32\,kW)(2.5\,h) = 10.8\,kWh$.

   • Cost: $(10.8\,kWh)(11.87\,\text{cents}/kWh) = 128.196\,\text{cents} \approx \$1.28$.

2. Tiered Billing (A):

   • Usage: $950\,kWh$ over 2 months. Rate: $11.87\,\text{cents}/kWh$.

   • Cost: $950 \times 11.87 = 11,276.5\,\text{cents} = \$112.77$.

3. Tiered Billing (B):

   • Usage: $1,850\,kWh$ over 2 months.

   • Step 1: $1,376\,kWh \times 11.87 = \$163.33$.

   • Step 2: $(1,850 - 1,376) = 474\,kWh \times 14.08 = \$66.74$.

   • Total: $\$163.33 + \$66.74 = \$230.07$.

4. Heating Water with Coil:

   • Given: $R = 10\,\Omega, V = 120\,V$, Water: $2\,L$ (mass $m = 2\,kg$), $T_{initial} = 20^{\circ}C, T_{final} = 100^{\circ}C$. Specific heat $c = 4180\,J/kgK$.

   • Step 1: Energy required ($Q$): $Q = mc\Delta T = (2\,kg)(4180)(100 - 20) = 668,800\,J$

   • Step 2: Power of coil: $P = \frac{V^2}{R} = \frac{(120)^2}{10} = 1440\,W$

   • Step 3: Time taken: $t = \frac{E}{P} = \frac{668,800\,J}{1440\,W} = 464\,s$

   • Step 4: Cost ($\$0.11/kWh$):

     • Energy in $kWh$: $E = 1.44\,kW \times (\frac{464\,s}{3600\,s/h}) = 0.1856\,kWh$.

     • Cost: $0.1856 \times 0.11 = \$0.02$.