Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied...

13
Applied Physics 04-03-08 Circuits

Transcript of Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied...

Page 1: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Applied Physics04-03-08Circuits

Page 2: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Announcements

• Amanda Bynes (22)

• The Sound, Hearing and Circuits 1 quizzes will go down Monday, April 14 at 11:55 pm.

• Homework Problems: Chapter 11: 2, 6, 8, 10, 14, 19, 22

Page 3: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Ohm’s Law

Ohm’s Law V = IR

R is resistance in Ohm’s (Ω).

V is the electric potential in Volts (V).

I is the current in Amperes (A).

Ω = VA

Page 4: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

How much current is produced when there is a 10 V potential difference across a 5 Ω resistance?

Question 1

How much resistance is present when a 10 V potential difference produces a 0.05 A current?

Question 2

How much potential difference is required to produce a 0.3 A current through a 15 Ω resistance?

Question 3

Page 5: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Electric Power

Power: P = VI = V2/R = I2R

Electric Potential:

V = EnergyCharge

Electric Current:

I = ChargeTime

V×I = (ChargeTime )Energy

Charge( ) (EnergyTime )=

Page 6: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

How much power is used by a circuit with a 10 V potential difference and a current of 0.25 A?

Question 4

How much power is used by a circuit with a 10 V potential difference and a resistance of 5 Ω?

Question 5

How much power is used by a current with a 0.25 A current and a resistance of 5 Ω?

Question 6

Page 7: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Drawing CircuitsI) Circuit Parts

+Battery

-

Resistor

Wire

II) Simple CircuitV

+-

RIII) A Circuit must make a loop.

Page 8: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Series ResistorsResistors that come one after another are in series.

To move around the circuit, you must pass through each of the series resistors.

V+-

R2 R1

Total Resistance: Rtot = R1 + R2Total Current: Itot = I1 = I2 = V/Rtot

Resistance AddsVoltage is sharedCurrent is same

Page 9: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Series Resistors

Q7) How much current is coming out of the battery?

V = 10 V+-

R2 = 5 Ω R1 = 5 Ω

Q8) How much current is passing through R1?

Q9) How much current is passing through R2?

A) 2.0 A B) 1.0 AC) 0.5 A D) 0.0 A

Page 10: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Series Resistors

Q10) How much potential difference is supplied by the battery?

V = 10 V+-

R2 = 5 Ω R1 = 5 Ω

Q11) How much potential is lost passing through R1?

A) 10 V B) 5 VC) 2 V D) 1 V

Q12) How much potential is lost passing through R2?

Page 11: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Parallel CircuitsResistors that come on different branches of a circuit are parallel.

To move around the circuit, you do not have to pass through each of the parallel resistors.

Total Resistance: 1/Rtot = 1/R1 + 1/R2Total Current: Itot = I1 + I2 = V/Rtot 1/Resistance Adds

Voltage is sameCurrent is shared

V+-

R2

R1

I1 = V/R1 and I2 = V/R2

Page 12: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Parallel Circuits

V = 10 V+-

R2 = 10 Ω

R1 = 5 ΩQ15) How much current is coming out of the battery?

Q14) How much current is passing through R1?

Q13) How much current is passing through R2?

A) 10.0 A B) 5.0 AC) 2.0 A D) 1.0 A

Page 13: Applied Physics - faculty.uca.edufaculty.uca.edu/carlf/PHYS_1405/Notes/04-03-08AP.pdf · Applied Physics 04-03-08 ... Ohm’s Law V = IR R is resistance in Ohm’s (Ω). V is the

Parallel Circuits

V = 10 V+-

R2 = 10 Ω

R1 = 5 Ω

Q17) How much potential is dropped across R1?

Q16) How much potential is dropped across R2?

A) 10 V B) 5 VC) 2 V D) 1 V