Circuits 101

I, V, R, and P

Today…

The electric current, I, is the rate of flow of charge Q through a given area in a given amount of time in an electric conductor.

IQ

t

Units: Coulomb/second = Ampere (A)

Current = charge time

 Andre M. Ampere  1775-1836

I, ELECTRIC CURRENT

; q

I q Itt

q = (6 A)(3 s) = 18 C

Recall that: 1 e- = 1.6 x 10-19 C, then convert:

-

20-19

1e18 C 18 C 1,125 x 10 electrons

1.6 x 10 C

In 3 s: 1.12 x 1020 electrons

Example 1. The electric current in a wire is 6 A. How many electrons flow past a given point in a time of 3 s? I = 6A

t = 3 se- = ?

Conventional Current vs Electron Flow theory

OR

Physics vs Chemistry

Which way does the current go?

:

?

Electron flow: Current flows in the direction an e-

moves; from – to +.

Conventional current: Current flows in the direction a +q moves; from + to –

++

--

+ -Electron flow

+ -+ -

e-

Conventional flow

+

Imagine a charged capacitor with Q = CV that is allowed to discharge

SOOOO. We will use conventional current as our preferred method for describing the flow of current in a circuit.

In a wire (a conducting metal), electrons are the only charged particles moving in an electrical current.

A source of electromotive force (emf) is a device that converts, chemical, mechanical, or other forms of energy into the electric energy necessary to maintain a continuous flow of electric charge.

The source of the emf in this case is chemical energy.

ELECTROMOTIVE FORCE

V

Example: Fred’s portable DVD player draws 0.600A of current. If new batteries supply 6400 C of charge until they run out, how long does the DVD player run?

IQ

t

I = 0.600 AQ= 6400 C

t = Q I

= 6400 C 0.600 C/s

= 10,666 sec

If the DVD player used 3V, how much energy does the battery use to run the player?

V = W qo

W = Vqo = (3V)(6400 C)

= 19,200 J

Resistance is an opposition to the flow of charge

Resistors

ResistanceR,

A resistor in a circuit is ANYTHING that opposes the flow of the current or uses voltage

This includes the wires of the circuit.

RESISTIVITY The resistance of a wire of uniform cross-sectional area is determined by:

The kind of materialThe lengthThe cross-sectional areaThe temperature

Rl

A

Where ρ is the resistivity of the material in Ω.m, l is the length in m, and A is the cross-sectional area in m2.

Digit – digit - multiplier 10X

4 band resistors

Orange – Black - Brown       

3 0 x 10Ω

300 Ω

More practice later!

Example: What is the resistance of a 20 m length of copper wire with a diameter of 0.8 mm?

ρ = 1.72x10-8 Ω.m l = 20 mr = d/2 = 4x10-4 m

A = πr2

= π(4x10-4)2

= 5.02x10-7 m2

Rl

A 1.72 10-8x

20

5 02 10 7. x= 0.685 Ω

"For a given resistor at a particular temperature, the current is directly proportional to the applied voltage."

RV

I

Units: Volt/Amperes = ohm (Ω)

Georg Simon Ohm

(1787-1854)

Ohm’s Law

Example: The voltage between the terminals of an electric heater is 80 V when there is a current of 6 A in the heater. What is the current if the voltage is increased to 120 V?

V1 = 80 V

I = 6 AV2 = 120 VR

V

I

80

6 = 13.3 Ω

IV

R

120

133.= 9 A

ELECTRIC POWERThe amount of work done in a given unit of time.

P = V I Units: volts x amperes = J/C x C/s = J/s = watts (W)

P = I2 R PV

R

2

P = V I

The faster the transfer of charge, the more power generated in the circuit.

P,

Example: A current of 6A flows through a resistance of 300 Ω for 1 hour. A) What is the power?

I = 6 AR = 300 Ωt = 1 hour

P = I2R = (6A)2(300Ω) = 10,800 W

b. How much heat is generated?

E = Pt = 10800J/s(3600s) = 3.89x107 J

This question asks, how many joules of electric energy are are converted into heat energy. To answer we need to remember that Power is Joules/sec, so to find the amount of energy we used, we take the power and multiply by time it runs in seconds

Example: The Elder’s keep their 40 W porch light on at night to welcome late night visitors. If the light in on from 6 pm to 7 am, and the Elder’s pay 0.1134 $/kwh, how much does it cost to run the light for a week?

= (?kW)(?hr) (0.1134 $/kWh)

7 days x 13 hrs x = 91 hr 1 day

40 W x 1kW = 0.040 kW 1000 W

= (0.040 kW)(91 hr) (0.1134 $/kWh)

Cost = E ($/kWh)

Cost = Pt ($/kWh)

= $ 0.41

Now how do we build circuits?

RESISTORS IN SERIES

To wire in SERIES means to make a circuit with a single pathway for the current.

RULES FOR RESISTORS IN SERIES

- In a series circuit, the current is the same at all points along the wire.  IT = I1 = I2 = I3

 - An equivalent resistance is the resistance of a single resistor that could replace all the resistors in a circuit. The single resistor would have the same current through it as the resistors it replaced.  RE= R1 + R2 + R3

 - In a series circuit, the sum of the voltage drops equal the voltage drop across the entire circuit.

VT = V1 + V2 + V3

Example: Two resistances of 2 Ω and 4 Ω respectively are connected in series. If the source of emf maintains a constant potential difference of 12 V,a. What is the current delivered to the external circuit?

Re = R1 + R2

= 2 Ω + 4 Ω = 6 Ω

IV

RTe

12

6= 2 A

b. What is the potential drop across each resistor?

V1 = I R1

= 2A(2 Ω) = 4 VV2 = I R2

= 2A(4 Ω) = 8 V

RESISTORS IN PARALLEL

To wire in PARALLEL means to make a circuit with multiple pathways for the current.

Rules for PARALLEL CIRCUITS

- In a parallel circuit, each resistor provides a new path for electrons to flow. The total current is the sum of the currents through each resistor.  IT = I1 + I2 + I3

- The equivalent resistance of a parallel circuit decreases as each new resistor is added.  

- The voltage drop across each branch is equal to the voltage of the source.

VT = V1 = V2 = V3

1 1 1 1

1 2 3R R R RE

Example: A 5Ω and 10Ω resistor are wired in parallel to 6V battery. A. What is their equivalent resistance?

R1= 5ΩR2 = 10ΩV= 6V

1 = 1 + 1RT 5 10

RT = 3.33 Ω

B. What is the current through each resistor?

IV

R = 6 V

5 Ω= 1.2 A

= 6 V 10 Ω

= 0.6 AIV

R

IV

R = 6 V

3.33Ω

IT= 1.80A

IT= 1.80A

Tonight and tomorrow night…Work problems # 1- 10

We will be watching…