Physics 212 Lecture 9, Slide 1 Physics 212 Lecture 9 Today's Concept: Electric Current Ohm’s Law & resistors Resistors in circuits Power in circuits

Physics 212 Lecture 9, Slide Physics 212 Lecture 9, Slide 11

Physics 212Physics 212Lecture 9Lecture 9

Today's Concept:Today's Concept:

Electric CurrentElectric Current

Ohm’s Law & resistorsOhm’s Law & resistorsResistors in circuitsResistors in circuits

Power in circuitsPower in circuits

Main Point 1

First, we defined the electric current in a conductor as the amount of charge that passes through a cross-section of the conductor per unit time (i.e., I = dq/dt). We developed a microscopic view of current in which the charge carriers are dissociated electrons in the conductor that move randomly at high speeds (~10^5 m/s), but when a potential difference is introduced across the conductor, the resultant electric field gives these electrons a non-zero average velocity which is the source of the electric current.


Main Point 2

Second, we introduced Ohm’s law that states that, for a wide range of materials, over a wide range of field strengths, the current density is proportional to the electric field that gives rise to it. The constant of proportionality is called the conductivity of the material. We used this constant to characterize the resistance of a resistor, obtaining the more common form of Ohm’s law in which the voltage across the resistor is equal to the product of the resistance and the current that flows through the resistor.


Main Point 3

Third, we obtained expressions for the equivalent resistance of two resistors connected either in series or in parallel. Namely, the equivalent resistance of two resistors in series is simply the sum of the individual resistances, while the inverse of the equivalent resistance of two resistors connected in parallel is equal to the sum of the individual inverse resistances.


Main Point 4

Fourth, we determined that the power, the time rate of change of the energy of a circuit component is always equal to the product of the voltage drop across the component and the current that flows through the component. In a simple circuit composed of a single resistor connected to the terminals of the battery, we found that all of the energy that is supplied by the battery is dissipated as heat in the resistor.



JJ = = EE

II AA

VV

R =R = L L AA

LL

II = = V/RV/Rsame assame as

wherewhere

Conductivity – high for good conductors.Conductivity – high for good conductors.

LL AA

resistivity – high for bad conductors.resistivity – high for bad conductors.

==



Checkpoint 1aCheckpoint 1a Checkpoint 1bCheckpoint 1b

Same current Same current through both through both

resistorsresistors

Compare voltages Compare voltages across resistorsacross resistors



Checkpoint Checkpoint 33The SAME amount of current I passes through three different resistors. R2 has twice the cross-

sectional area and the same length as R1, and R3 is three times as long as R1 but has the same cross-sectional area as R1.

In which case is the CURRENT DENSITY through the resistor the smallest?A. Case 1 B. Case 2 C. Case 3



Checkpoint 2aCheckpoint 2a

Compare the current through R2 with the current through R3:A. I2 > I3 B. I2 = I3 C. I2 < I3

Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R1= R2 = R3 = R.



Checkpoint 2bCheckpoint 2b

RR11 = R = R22 = R = R33 = R = R

A. A. II11/I/I22 = 1/2 = 1/2

B. B. II11/I/I22 = 1/3 = 1/3

C. C. II11/I/I22 = 1 = 1

D. D. II11/I/I22 = 2 = 2

E. E. II11/I/I22 = 3 = 3



RR11 = R = R22 = R = R33 = R = R

Checkpoint 2cCheckpoint 2c

Compare the voltage Compare the voltage across Racross R22 with the with the voltage across Rvoltage across R33

VV22 > V > V33

VV22 = V = V3 3 = V= V

VV22 = V = V3 3 < V< V

VV22 < V < V33

AA

BB

CC

DD



RR11 = R = R22 = R = R33 = R = R

Checkpoint 2dCheckpoint 2d

Compare the voltage Compare the voltage across Racross R11 with the voltage with the voltage across Racross R22

VV11 = V = V2 2 = V= V

VV11 = ½ V = ½ V2 2 = V= V

VV11 = 2V = 2V2 2 = V= V

VV11 = ½ V = ½ V2 2 = 1/5 V= 1/5 V

VV11 = ½ V = ½ V2 2 = ½ V= ½ V

AA

BB

CC

DD

EE



Voltage

Current

Resistance

Series Parallel

Resistor Summary

Different for each resistor.Vtotal = V1 + V2

IncreasesReq = R1 + R2

Same for each resistorItotal = I1 = I2

Same for each resistor.Vtotal = V1 = V2

Decreases1/Req = 1/R1 + 1/R2

WiringEach resistor on the same wire.

Each resistor on a different wire.

Different for each resistorItotal = I1 + I2

R1 R2

R1

R2


CalculationCalculationIn the circuit shown: V = 18V,

R1 = 1R2 = 2R3 = 3andR4 = 4

What is V2, the voltage across R2?

• Conceptual Analysis: – Ohm’s Law: when current I flows through resistance R, the

potential drop V is given by: V = IR.– Resistances are combined in series and parallel combinations

• Rseries = Ra + Rb

• (1/Rparallel) = (1/Ra) + (1/Rb)• Strategic Analysis

– Combine resistances to form equivalent resistances– Evaluate voltages or currents from Ohm’s Law– Expand circuit back using knowledge of voltages and currents

V

R1 R2

R4

R3


Documents

Physics 212 Lecture 9, Slide 1 Physics 212 Lecture 9 Today's Concept: Electric Current Ohm’s Law & resistors Resistors in circuits Power in circuits