Transcript R 3
Internal Resistance in batteries Batteries A battery supplies nearly constant voltage NOT current Notice headlights dim when you start the car. Headlights draw some current; starter motor draws more Inference: voltage from batteries is not constant, but depends somehow on current Internal Resistance To supply voltage, charged particles move within electrolyte. Hindrance to the motion of charged particles = internal resistance. ππ π’ππππππ π‘π πππππ’ππ‘ = π β πΌπ Batteryβs rated emf Current through circuit Internal resistance is typically abbreviated with lower-case r Example (part A) A device with a 65-ο resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 ο. Calculate the current in the circuit. ππ π’ππππππ π‘π πππππ’ππ‘ = π β πΌπ where ππ‘π πππππ’ππ‘ = πΌπ So, πΌπ = π β πΌπ So, πΌ(π + π) = π So πΌ = π π +π = 12.0 π 65 ο+0.5 ο = 0.18 π΄ Example (part B) A device with a 65-ο resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 ο. Calculate the terminal voltage. ππ‘π πππππ’ππ‘ = π β πΌπ = 12 π β 0.18 π΄ 0.5 ο = 11.9 π Example (part C) A device with a 65-ο resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 ο. Calculate the power dissipated by the device. π = πΌπ = πΌ2 π = 0.18 π΄ 2 (65 ο) = 2.18 π Example (part D) A device with a 65-ο resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 ο. Calculate the power dissipated by the internal resistance in the battery. π = πΌπ = πΌ2 π = 0.18 π΄ 2 (0.5 ο) = 0.02 π Batteries get warm! And get warmer with larger currents. Power from a circuit Purpose The purpose of creating circuits is to provide power to components in that circuit, i.e., the things we want to do something useful β’ How much voltage? β’ How much resistance? β’ How much current? β’ How much power? Simplified home circuit Devices have different power requirements. β’ Transformers convert the potential to the appropriate voltage. β’ Resistors convert the current to the appropriate levels; devices themselves can be modeled as resistors. β’ Capacitors prevent surges and, with resistors, control timing. Adapters Understanding resistance in series circuit Charge is conserved. When electricity travels through components in series, πΌ = constant Same current but ο± resistance ο° ο± voltage drop ο° ο± power Simple series circuit Charge is conserved. π = π πΌ, so V3ο = (4 A) (3 ο) = 12 V V6ο = (4 A) (6 ο) = 24 V 4 A through 3 ο resistor, ο 4 A through 6 ο resistor (and everywhere else) π = ππΌ, so P3ο = (12 V)(4 A) = 48 W P6ο = (24 V) (4 A) = 96 W voltage Voltage drops around a circuit Energy is conserved. The sum of changes in potential energy throughout the circuit must add to zero. π½πππππππ = π½ππππππππ π + π½ππππππππ π + π½ππππππππ π + β¦ Kirchoff Loop Rule ππππ‘π‘πππ¦ = ππππ ππ π‘ππ 1 + ππππ ππ π‘ππ 2 + ππππ ππ π‘ππ 3 + β¦ βπ = 0 ππππ Sometimes called Kirchoff Loop Rule First explored and articulated by a German physicist, Gustav Kirchoff (1824-1887) Implications ππ£πππ‘πππ π ππ’πππ = ππππ ππ π‘ππ 1 + ππππ ππ π‘ππ 2 + ππππ ππ π‘ππ 3 + β¦ πΌπ π£πππ‘πππ π ππ’πππ = πΌπ 1 + πΌπ 2 + πΌπ 3 + β― When arranged in series, πΌ is constant, so π πππ’ππ£πππππ‘ = π 1 + π 2 + π 3 + β― Summary of resistors in series In a circuit with resistors in series, β’ Current is constant, πΌ = ππππ π‘πππ‘ β’ Overall resistance is the sum of resistance of individual resistors, π ππ = π 1 + π 2 + π 3 + β― β’ Voltage drop across each resistor is proportional to current and resistance of individual resistors. tip: π1 = π 1πΌ, π2 = π 2πΌ, Pro π = π πΌ 3 3 Memorize that current is constant in series! (a) Youβll understand the physics better (b) You can derive the rest of these equations if you also remember that V = RI Example A 10,000-ο resistor is placed in series with a 100-ο resistor. The current in the 10,000-ο resistor is 10 A. If the resistors are swapped, how much current flows through the 100-ο resistor? A. Less than 10 A B. 10 A C. More than 10 A D. Not enough information to determine Example (part A) Two devices, each with 100 ο resistance, are connected in series to a 24.0 V battery. What is the equivalent resistance? π ππ ππ π πππππ = π 1 + π 2 π ππ ππ π πππππ = 100 ο + 100 ο = 200ο Example (part B) Two devices, each with 100 ο resistance, are connected in series to a 24.0 V battery. What is the current through the resistors? ππ‘ππ‘ππ = π1 + π2 where π1 = πΌπ 1 and π2 = πΌπ 2 ππ‘ππ‘ππ = πΌπ 1 + πΌπ 2 = πΌ(π 1 + π 2) So, πΌ = ππ‘ππ‘ππ π 1+π 2 = 24.0 π 100 ο + 100 ο πΌ = 0.120 π΄ Simple parallel circuit The potential difference is the same on each branch of the parallel circuit. Current at junctions Charged particles make up current. Particles may transfer energy, but are not created or destroyed. The amount of charge entering any branch of a circuit must equal the amount of charge leaving the circuit. Kirchoff Junction Rule πΌ= ππ πΌ Sometimes called Kirchoff Junction Rule ππ’π‘ πΌπ‘ππ‘ππ = πΌ1 + πΌ2 + πΌ3 + β― π π π‘ππ‘ππ π π π = + + +β― π 1 π 2 π 3 where V is the same across each resistor Resistance in parallel branches π π π‘ππ‘ππ π π π = + + +β― π 1 π 2 π 3 1 1 1 1 = + + +β― π π‘ππ‘ππ π 1 π 2 π 3 Summary In a circuit with resistors in parallel, β’ Voltage is the same, π1 = π2 = π3 = β― β’ Overall resistance is 1 π ππ = 1 π 1 + 1 π 2 1 + π 3 +β― β’ The current through each branch of circuit is proportional to current and inversely proportional to the resistance of individual resistors. π πΌ1 = , π 1 π πΌ2 = , π 2 π πΌ3 = , π 3 Example (part A) Two devices, each with 100 ο resistance, are connected in parallel to a 24.0 V battery. What is the equivalent resistance? π ππ ππ ππππππππ = π ππ ππ π πππππ = 1 1 + 1 π 1 π 2 1 1 1 + 100 ο 100 ο = 50 ο Example (part B) Two devices, each with 100 ο resistance, are connected in parallel to a 24.0 V battery. What is the current through the resistors? πΌπ‘ππ‘ππ = πΌπ‘ππ‘ππ = π πΌ1 + πΌ2 where πΌ1 = , πΌ2 π 1 π π 24 π 24 π + = + π 1 π 2 100 ο 100 ο πΌπ‘ππ‘ππ = 0.48 π΄ = π π 2 Compare π ππ = 200 ο πΌ = 0.12 π΄ ππ 1 = 12 π π = 2.9 π π ππ = 50 ο πΌ = 0.48 π΄ ππ 1 = 24 π π = 11.5 π What is the overall resistance of this set of resistors? Think about it. Commit your answer to paper. Move this box to check the answer. What is the overall resistance of this set of resistors? Think about it. Commit your answer to paper. Move this box to check the answer. What is the overall resistance of this set of resistors? Think about it. Commit your answer to paper. Move this box to check the answer. VIRP table By combining your knowledge of Ohm's Law, Kirchoff's Junction Law, and Kirchoff's Loop Law, you can use this table to solve for the details of any circuit. V I R P Voltage (V) Current (A) resistance (ο) Power (W) R1 R2 R3 Total Adapted from http://www.aplusphysics.com/courses/honors/circuits/meters.html Problem-Solving V Step 1: Identify your givens I Voltage Current (V) (A) R1 R2 R3 R4 Total R resistance (ο) 10 10 16 4 60 P Power (W) Problem-Solving In this example, it makes sense to next calculate overall resistance. 1V I π ππ = Voltage 1Current 1 + 10 ο + 10 ο (V)16ο +(A) 4ο π ππ = 10Rο1 R2 R3 R4 Total P R resistance (ο) 10 10 16 4 60 Power (W) Problem-Solving Calculate overall current: V I π 60 π πΌ = = = 6 Voltage π΄ Current π 10 ο (V) (A) R1 Calculate overall power π = πΌπ = (60 π) (6 π΄) = 360 π R2 R3 R4 Total 60 P R resistance (ο) 10 10 16 4 10 Power (W) Problem-Solving V I Voltage Current (V) (A) Next, calculate current through each R1 branch π 60 π R πΌ10 ο + 10 ο = = = 32 π΄ π 10 ο + 10 ο 20 ο R3 π 60 π πΌ16 ο + 4 ο = = =R 3 4π΄ π 16 ο + 4 ο 20 ο Total 60 R resistance (ο) P Power (W) 10 10 16 This result is a function of this particular 4 circuit. Current through different 6 branches10 360 is NOT necessarily the same. Problem-Solving Calculate the voltage drop across each resistor: π = π πΌ π10ο = 10 ο 3 π΄ = 30 π π16ο = 16 ο 3 π΄ = 48 π π4ο = 4 ο 3 π΄ = 12 π R1 R2 R3 R4 Total V I Voltage Current (V) (A) 30 30 48 12 60 3 3 3 3 6 P R resistance (ο) Power (W) Confirm 10each branch sums to the appropriate voltage. 10+ 30 π = 60 π 30 π 48 π 16+ 12 π = 60 π 4 10 360 Problem-Solving Calculate the power dissipated by each resistor: π = ππΌ π10ο = 30Vπ 3 π΄I = 90 πR π16ο = 48Voltage π 3 π΄Current = 154 π resistance (ο) (V) (A) π4ο = 12 π 3 π΄ = 36 π R1 30 3 R2 30 3 RConfirm 48 dissipated 3 power by 3 resistors sums to Rindividual 12 3 4 power dissipated by overall Total 6 circuit. 60 90 π + 90 π + 154 π + 36 π = 360 π 10 10 16 4 10 P Power (W) 90 90 154 36 360 Problem-Solving R1 R2 R3 R4 Total V I R P Voltage (V) Current (A) resistance (ο) Power (W) 30 30 48 12 60 3 3 3 3 6 10 10 16 4 10 90 90 154 36 360 Example V I Voltage Current (V) (A) R1 R2 R3 Total R resistance (ο) P Power (W) Use your knowledge of Ohmβs Law and Kirchoffβs Laws to fill Youβre really Answers arenβt stuck? here.I encourage Keep trying. you to keep trying. in the VIRP table. Step 9 Example Step 5 π = πΌπ = (0.01π΄)(5 π) π = 0. 05 π Entire current passes through 400 ο resistor. Therefore, π@400ο = πΌπ = (0.017π΄)(400 ) Step 7 π = πΌπ where π ππ = 1 1 500 ο + 1 700 ο π = 0.017 π΄ 290 ο π = 5.0 π V I Step 8 πΌ= R1 7.0 R2 5.0 R3 5.0 Total 12.0 5.0 π = 7.1 π₯ 10β3 π΄ 700 ο 5.0 π πΌ= = 0.01 π΄ 500 ο 0.017 0.01 0.007 0.017 Step 3 πΌ= π 12.0π = π 690 ο πΌ = 0.017 π΄ π = πΌπ π = 0. 0035 π Voltage Current (V) (A) π@400ο = 7.0 V Step 6 π = πΌπ = (0.0071π΄)(5 π) = (0.017π΄)(6.8 π) R π = 0. 2 π Power (W) resistance (ο) 400 500 700 690 0.12 0.05 0.035 0.2 Step 2 π ππ = 400 ο + 1 1 1 + 500 ο 700 ο π ππ = 690 ο P Step 4 π = πΌπ = (0.017π΄)(12 π) π = 0. 2 π Example V I Voltage Current (V) (A) R1 7.0 R2 5.0 R3 5.0 Total 12.0 .017 0.01 0.007 0.017 P R resistance (ο) 400 500 700 690 Step 10 Confirm charge is conserved through branches. Confirm potential drop = potential gain Confirm power dissipated through individual resistors = total power dissipated. (Differences here due to rounding) Power (W) 0.12 0.05 0.035 0.2 Circuit diagram conventions