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Transcript
Current, Resistance and Power Battery Negative Electrode - electrolyte + Positive Electrode Battery Negative Electrode - electrolyte + Positive Electrode Current Simple flow of charge See Active Figure 27.09 I (Note Convention) - dQ I dt Current MKS unit - Ampere Current Charge carriers + + + + + A + n - concentration of charges per unit volume vd – drift velocity A – cross sectional area of conducting wire q – charges carried by each particle After some time t, the particles will pass beyond a particular point on the wire + A + + + + + l vd t The volume contains the passing charges can be found with vd, A and t Vol Al Avd t Volume of passing charges + A + + + + + l vd t Q Nq Q (nVol )q + + + + Q nAvd tq Q I nAvd q t + + Current Density and Ohm’s Law I J A Current Density Current per unit Area Current Density and Ohm’s Law J E Ohm’s Law - conductivity I J dA A more familiar form J E I V E A l l V I A 1 Resistivity V IR Ohm’s Law: Macroscopic form where R l A Resistance Volt/amp= W MKS Unit Note Dependencies • If you double the area (ie. Adding an addition wire) the effective resistance halves • If you add the wire to the length the effective resistance doubles • The resistivity is an intrinsic property of the material the resistor is made of. If you change material keeping physical geometry the same, the resistance changes I Ohmic (or linear) device V Slope = 1/R I Non-Ohmic (or nonlinear) device V Microscopic View of Conductor Q I nAvd q t I 5 Amps vd 0.37mm / s 28 3 -6 2 -19 nAq 8.48 10 atoms / m 10 m 1.6 10 C copper Cross sectional area J E Electron Charge acceleration qE J nqvd nq E m Time between collisions e2 n me See Active Figure 27.09 Independent of Electric Field: Ohmic But can depend on conditions which effect , such as temperature Resistivity vs. Temperature (T) – characteristic of material T metallic T semi-conductors insulators 0 T0 T 0 (1 (T - T0 )) 0 – resistivity at T0 – thermal coefficient of resistivity Superconductors • Many materials will below a specific characteristic temperature, Tc, have a pronounced decrease in resistivity. Power Battery – “works” to push current through circuit I Powersource = VI V V – Potential Source I – Current sent from source through circuit Thermal energy dissipated through resistors R I dW dU dq V P dt dt dt 2 V P IV I 2 R R Voltage drop across resistor Rate of Thermal Energy dissipation through Resistor Example Problem: Suppose we wanted to design a small heater for your to work before your car warmed up. We want 500Watts using the 12V of your car battery. How much Nichrome wire with a crossectional area of 0.1 cm2 do we need? 2 V P IV I R R 2 12V 500 watts R R 0.29W 2 R A 100W cm 0.29 W 0.1cm 2 290 cm Battery (Source) e r e- actual potential difference between electrodes of battery (EMF) r – internal resistance of battery Battery (Source) e r I R By attaching the battery to a circuit including a load resistor R, the current drawn through the battery will effect the actual potential difference in the battery Kirchoff’s Voltage Loop Theorem • The algebraic sum of the changes in electric potential encountered in a complete traversal of the circuit must be zero. • A circuit is closed path through which current (electrons) may be forced to move through circuit elements (resistors). V = e - Ir Battery voltage terminal to terminal e r I I R Jumping from the negative to the positive end of the battery, the potential increases by e, but after going through the resistor, the potential drops by IR To find the current… e r I I R e - Ir - IR 0 e I (r R) Kirchoff’s Voltage Loop Resistors in Series and Parallel: Equivalent Resistance Resistors in Series R1 I e R2 I R3 e - IR1 - IR2 - IR3 0 e - I ( R1 R2 R3 ) 0 Resistors in Series I e Rseries I e - IRseries 0 Rseries ( R1 R2 R3 ) Kirchoff’s Junction Theorem • At any junction (point where current can split) the algebraic sum of the currents into and out of the wires of the junction must add to zero. • By convention the current into a junction is positive and the current out of a junction is negative. Resistors in Parallel I e R1 I R2 R3 Resistors in Parallel I e I1 I2 R1 I3 R2 I I I1 I 2 I 3 R3 Resistors in Parallel I e I1 I2 R1 I3 R2 I e - I1R1 0 I1 e R1 R3 Resistors in Parallel I e I1 I2 R1 I3 R2 I e - I 2 R2 0 I 2 e R2 R3 Resistors in Parallel I e I1 I2 R1 I3 R2 I e - I 3 R3 0 I 3 e R3 R3 Resistors in Parallel I e I Rparallel I e - IR parallel 0 I e Rparallel Resistors in Parallel I I1 I 2 I 3 e R parallel 1 R parallel e R1 e R2 e R3 1 1 1 R1 R2 R3 Solve for the currents going through each of the resistors by circuit reduction (equivalent resistance) 12 W 42 V 1W 2W 2W 4W Break circuit down into series and parallel resistors Currents in the various branches 12 W I I2 42 V I1 1W 2W 2W 4W 12 W I I2 42 V I1 1W 2W 2W 4W Find equivalent resistance for the Series Resistors 12 W I I2 42 V 3W I1 Find Equivalent Parallel Resistance 6W 12 W I 42 V 2W Find Equivalent Series Resistance I 42 V 14 W Find Equivalent Series Resistance I 42 V 14 W 42 - 14 I 0 I 3A 12 W I I2 42 V I1 1W 2W 2W 4W 42 - 12 I - 1I1 - 2 I1 0 3I1 42 - 12 I 6V I1 2 A 12 W I I2 42 V I1 1W 2W 2W 4W I I1 I 2 I 2 I - I1 1 A Kirchoff’s Analysis e1 - I1 (6W) e 2 I 2 (4W) 0 e1 - I1 (6W) - I 3 (2W) 0 I1 I 2 - I 3 0 Solve simultaneously for the unknown currents