Final Exam Review Current, Power, Voltage, Resistance Ohm’s Law Series & Parallel Circuits Electromagnetic Induction.

Final Exam Review

Current, Power, Voltage, ResistanceOhm’s Law

Series & Parallel CircuitsElectromagnetic Induction

Calculating a Power Bill

• Formula: $ = rPt• Bill = Rate * Power (kW) * time (h)

• For this application ONLY, Power is in kilowatts, and time is in hours.

Example

• At 0.007 $/kWh, calculate the cost of running a 1000W microwave for 30 minutes.

• $ = rPt• $ = (0.007 $/kWh)(1kW)(0.5h)• $ = $0.00350

=

Practice

• Pg 623 Section Review• # 4.

Current, Power, Voltage

• 3 main equations:• Q = It• Charge (C) = Current (A) * time (s)

• V = IR• Voltage (V) = Current (A) * Resistance (Ω)

• P = IV• Power (W) = Current (A) * Voltage (V)

Circuit Types• Series Circuits: Circuits with only one path for

electricity to flow through.

Circuit Types

• Parallel Circuit: Circuit with two or more loops.

Properties of Circuit Types

• Series Parallel• Itot = I1 = I2 = I3 =… Itot = I1 + I2 + I3 + …

• Vtot = V1 + V2 + V3 +… Vtot = V1 = V2 = V3 =…

• Rtot = R1 + R2 + R3 +… 1/Rtot = 1/R1+1/R2+…

Example- Series Circuit

• A 9.0V battery is connected to four light bulbs in series, with resistances of 4.0 Ω, 5.0Ω, 7.0Ω, and 2.0Ω. Find the equivalent resistance for the circuit and the current in the circuit.

Example

• A 9.0V battery is connected to four light bulbs in series, with resistances of 4.0 Ω, 5.0Ω, 7.0Ω, and 2.0Ω. Find the equivalent resistance for the circuit and the current in the circuit.

• Rtot = R1 + R2 + R3 + R4

Example

• A 9.0V battery is connected to four light bulbs in series, with resistances of 4.0 Ω, 5.0Ω, 7.0Ω, and 2.0Ω. Find the equivalent resistance for the circuit and the current in the circuit.

• Rtot = R1 + R2 + R3 + R4

• Rtot = 4.0 + 5.0 + 7.0 + 2.0 = 18.0Ω

Example

• A 9.0V battery is connected to four light bulbs in series, with resistances of 4.0 Ω, 5.0Ω, 7.0Ω, and 2.0Ω. Find the equivalent resistance for the circuit and the current in the circuit.

• Rtot = R1 + R2 + R3 + R4

• Rtot = 4.0 + 5.0 + 7.0 + 2.0 = 18.0Ω

• V=IRtot

Example

• A 9.0V battery is connected to four light bulbs in series, with resistances of 4.0 Ω, 5.0Ω, 7.0Ω, and 2.0Ω. Find the equivalent resistance for the circuit and the current in the circuit.

• Rtot = R1 + R2 + R3 + R4

• Rtot = 4.0 + 5.0 + 7.0 + 2.0 = 18.0Ω

• V=IRtot

• 9.0V = I (18.0Ω)• I = 0.5A

Practice

• Pg 650 Practice A• 1 & 2

Example- Parallel Circuit

• A 9.0V battery is connected to two resistors of 2.0Ω and 4.0Ω in parallel. Find the equivalent resistance for the circuit and the total current in the circuit.

Example

• A 9.0V battery is connected to two resistors of 2.0Ω and 4.0Ω in parallel. Find the equivalent resistance for the circuit and the total current in the circuit.

• 1/Rtot = 1/R1 + 1/R2

Example

• A 9.0V battery is connected to two resistors of 2.0Ω and 4.0Ω in parallel. Find the equivalent resistance for the circuit and the total current in the circuit.

• 1/Rtot = 1/R1 + 1/R2

• 1/Rtot = 1/2.0 + 1/4.0 = .75

• Rtot = 1.33Ω

Example

• A 9.0V battery is connected to two resistors of 2.0Ω and 4.0Ω in parallel. Find the equivalent resistance for the circuit and the total current in the circuit.

• 1/Rtot = 1/R1 + 1/R2

• 1/Rtot = 1/2.0 + 1/4.0 = .75

• Rtot = 1.33Ω

• V= IRtot

Example

• A 9.0V battery is connected to two resistors of 2.0Ω and 4.0Ω in parallel. Find the equivalent resistance for the circuit and the total current in the circuit.

• 1/Rtot = 1/R1 + 1/R2

• 1/Rtot = 1/2.0 + 1/4.0 = .75

• Rtot = 1.33Ω

• V= IRtot

• 9.0 = I (1.33Ω) I = 6.92A

Practice

• Pg 655- Practice B• #3