Physics 106 Lesson 11 Series Parallel Circuits I

Physics 106 Lesson #11 Series & Parallel Circuits I Dr. Andrew Tomasch 2405 Randall Lab atomasch@umich. edu

Review: Cells, Batteries & Current • When connected to a circuit, the potential difference between the cell or battery terminals creates an electric force on the charges in the conductor causing them to move and establishing an electric current • The conventional current is from regions of higher potential to regions of lower potential, positive to negative Conventional current

Review: Good and Bad Wiring • Good: • Bad: Broken! • Ugly: Short! No device! =

Review: Resistors are used in appliances to convert electrical energy into thermal energy (heat) or light Toaster Space Heater Light Bulb When an extension cord is used with a space heater, the cord must have a resistance that is sufficiently small to prevent overheating of the cord Stove Heating Element

Review: Series Resistors • For resistors R 1 & R 2 connected in series (sequentially), the current i passing through each resistor must be the same • The voltages across R 1 & R 2 must add up to V (Loop Rule) V +

Review: Parallel Resistors • Resistors R 1 & R 2 connected in parallel have the same potential difference (voltage) V across them • Charge must be conserved, so the currents I 1 & I 2 flowing through the two resistors must add up to the total current I leaving the battery (continuity for electric current!) V +

Review: Power • Power has a precise definition in physics: Power is the rate at which work is done or how much work is done per unit of time Units: J/s Watt (W) James Watt invented the steam engine In an electric circuit power is the product of current and voltage:

Concept Test #1 The label on a car battery proudly claims that the battery will provide “ 250 Ampere hours”. This rating describes the amount of what quantity that can be drawn from the battery? 1) 2) 3) 4) 5) Charge Current Voltage Power Energy current =charge/time à current × time = charge à 1 Ampere-hour = 1 Coulomb/second x 3600 s/hr = 3600 Coulombs of charge

Concept Test #2 How much energy is stored in a 2 V battery with a capacity of 1 Ampere-hour? 1) 2) 3) 4) 5) 3, 600 Joules 1, 800 Joules 7, 200 Joules 10, 000 Joules 5, 000 Joules Energy = charge (C) x potential (V) à 1 Ampere-hour = 1 Coulomb/second x 3600 s/hr = 3600 Coulombs of charge à 3600 Coulombs x 2 V = 7, 200 Joules of energy

Concept Test #3 What is the average power delivered when I discharge an 8. 4 V battery with a capacity of 2 A-hr in 10 minutes? 1) 3, 600 Watts 2) 1, 800 Watts 3) 7, 200 Watts 4) 100 Watts Power = energy/time = charge (C) x potential (V)/time à 2 Ampere-hour = 2 Coulomb/second x 3600 s/hr = 7, 200 Coulombs of charge à 7, 200 Coulombs x 8. 4 V = 60, 480 Joules of energy à 10 min = 600 seconds à 60, 480 Joules/600 seconds ≈ 100 Watts of power