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Notes 33 - Topics 5 & 6 - Electricity & Magnetism *

Notes 33 - Topics 5 & 6 - Electricity & Magnetism * - ------------------------------------------------------------------------ 5 .1.11 Solve problems dealing with I, V, and R Sa mple Problem 1: A 10.0 V cell (neglect r ) is connected to a

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Notes 33 - Topics 5 & 6 - Electricity & Magnetism *

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  1. Notes 33 - Topics 5 & 6 - Electricity & Magnetism* ------------------------------------------------------------------------- 5.1.11 Solve problems dealing with I, V, and R Sample Problem 1: A 10.0 V cell (neglect r) is connected to a 14.0 Ω resistor. What current should flow through the resistor? Given: Unknown: Equation:

  2. Sample Problem 2: A 20.0 Ω resistor has 1.75 A of current flowing through it. What is the terminal voltage of the battery? Given: Unknown: Equation:

  3. Sample Problem 3: A 9.0 V battery (neglect r) is connected to a resistor. If the current passing through the resistor is 1.0 A, what is the resistance of the resistor? Given: Unknown: Equation:

  4. 5.1.10 Electrical Power Dissipation in Resistors; - When current passes through a resistor, the temperature of the resistor increases and energy is released, in the form of heat, into the environment each second; - The heat energy released per second (Js-1) is the electrical power dissipated (given off) by that resistor; - Electrical Work = charge x Voltage; We = q V = (C) (J C-1) = J ;

  5. - Electrical Power = Electrical Work / time; Pe = We / Δt = q V / Δt = J s-1 = watts = W ; ...but... q V / Δt = V (q / t) = V I ... so, Pe = I V = watts = W ; - From Ohm’s Law: V = I R; since Pe = I V, then Pe = I (I R) = I2 R* ; - From Ohm’s Law: I = V / R; since Pe = I V, then Pe = (V / R) V = V2 / R ; To summarize: Pe = I V = I2 R * = V2/ R (*Of the 3 forms, this is the most commonly used)

  6. Sample Problem 4: A 9.0 V battery (neglect r) is connected to a resistor. If the current passing through the resistor is 1.0 A, what is the resistance of the resistor? Given: Unknown: Equation:

  7. Sample Problem 5: From Sample Problem 4, show 3 ways to determine the power dissipated in the resistor. Given: Unknown: Equation:

  8. 5.1.2 Electric Potential Difference in a Conductor Electric Potential Difference (V) can be defined in 2 ways... A. Ve is the work done on a charge to move it from one potential to another in an electric field, or, the change in electric potential energy of the charge as it moves (5.1.1): Ve = We / q = ΔPEe / q B. The Vein a conductor carrying a current is the power dissipated per unit current moving between two points: Ve = P / I

  9. 5.2.1 Electromotive Force (emf or ε )... - is not a force at all but had the bad luck of being discovered and named before the concept of potential difference was known; - is the electric potential difference developed by a battery or a generator when no current is flowing (ie., the “shelf-voltage” of a battery); - is used interchangeably with electric potential difference and voltage; - is the energy supplied per charge while Ve is the energy dissipated per charge; - is symbolized by ε ... the words are rarely used

  10. 5.2.2 Internal Resistance (r)- resistance to the flow of electrons by the electrolyte of a battery (ie., resistance opposing the motion of the electrons inside the battery itself); • Shelf Voltage ( ε ) - the actual emf of a battery when it is NOT supplying current to a circuit: ε = VT + I r • Terminal Voltage (VT) - the voltage of a battery or generator when a resistance is present in a current carrying circuit; - Terminal Voltage equals emf (shelf voltage) minus current times internal resistance: VT = ε - I r

  11. Sample Problem 6: A cell has an internal resistance of 1.80 Ω. When a 14.0 Ω resistor is connected to the cell, the electric potential difference across the resistor is 1.40 V and the current through the resistor is 0.600 A. What is the emf of the battery? Given: Unknown: Equation:

  12. Sample Problem 7: A single cell “D” battery connected to a circuit with a 4.0 ohm resistance produces a terminal voltage of 1.30 V. The same “D” cell is placed in another circuit with a resistance of 12 ohm and the new terminal voltage is found to be 1.45 V. Find the internal resistance of the cell and the emf of the cell. Given: Unknown: Equation:

  13. Sample Problem 8: A 1.0 V emf photoelectric cell produces a 0.75 A current through a resistance of 1.0 ohm. What is the internal resistance of this P. E. cell? Given: Unknown: Equation:

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