## Electricty and Magnetism, Tutorial 3

1) Dielectric: field within dielectric is smaller than "expected" - polarized molecules slightly cancel out the field. Use å rather than å0 in all formulae and things will work out (probably not always, but in every case we have and are likely to look at).

2) Electric Field Density - U/V = 1/2 eps0E2 - Although this is most easily deriver from capacitor theory, this result holds true whenever there is an electric field. It shows the reality of the electric field. Important not to double count energy - either count the energy as being stored in the charges, or stored in the fields - not both.

3) Batteries - Chemistry in batteries causes voltage (basically gives electrons a kick around the circuit).

4) Current - I=DeltaQ/Deltat – the amount of charge flowing past a point in unit time.

5) Amps - Coulombs per second

6) Conservation of charge - "Kirchoff's Current Node Law" - Current into a node of wires equals the current out.

7) Current defined in terms of the movement of (usually fictional) positive charges

i. (+) -> (+) -> I ->

ii. (-) -> (-) -> <-I

8) Resistors - Ohms Law: V=IR, R=ñL/A

9) Power dissipation in a resistor (or similar element, like heater) - P=VI=V2/R=I2R

10) Alternating Current - Time dependence: V=V0sin(ùt+ö). V0: Peak Voltage. RMS (Root Mean Squared) Voltage: VRMS=V0/1.41 (only for sine wave voltage – other, more complicated, voltage forms have different RMS definitions).

11) P=VRMSIRMS=VRMS2/R=IRMS2R

12) Resistors in Series - R=R1+R2+…

13) Resistors in Parallel - R-1=R1-1+R2-1+…

14) Temperature dependence - ñ[T0+epsT]=ñ0(1+áÄT)

## Example Problems

A 1.6 Ohm bulb is connected to a 9V battery. How many electrons flow through it per second?

A) Bulbs can be treated like resistors - their relatively low resistance leads to a high current and lots of heat so they glow. However, because they heat up so much, their resistance varies considerably with the current. We'll assume that the resistance given is valid for the current we will calculate.

I = 9/1.6 A

I = 5.6C/s

I = 5.6C/s/(1.6x10-19 C/e)

I = 3.5x1019 electrons/second

## Resistance, Heat and Efficiency

Q) A heating element is used to increase the temperature of 150mL of water from 5oC to 95oC in 5 minutes. If the heater is 60% efficient and is powered by a 12V battery, what is its resitance?

A) The heat required to raise the temperature of a lump of matter is given by:

ΔQ= mass * ΔT * Specific Heat
(Q is the customary symbol for heat â€“ in this case it is not charge. Sorry, there are only 26 letters and many, many more quantities we need to talk about)

In this case,

ΔQ = .15kg * 90oC * 4186J/(kgoC)

ΔQ = 56.5kJ

Because the heater is only 60% efficient, for every joule of electrical energy put in, only .6J of heat energy makes it in to the water.
ΔE=56.5kJ/.6

ΔE=94.2kJ

If this energy flows in evenly over the 300s, then the power is:
P=94.2kJ/300s=314W

P=VI

I = 26A

R=V/I = 45Ohms

Room Heater Power

## Power Requirements

Q) A heater is used to keep a room at 20oC. Every half hour the air in the 62m3 is swapped for new air at 5oC (in reality this swap happens smoothly over the course of the half hour so the temperature remains stable). Additionally, heat is lost through the walls of the room at a rate of 850kJ/hour.

What is the minimum power required of the heater?

A) We can calculate the heat energy required to keep the room hot over the course of an hour: There are two separate heating requirements â€“ heating the new air and replacing the heat lost.

1) 5o -> 20o in 30mins

62m3 of air has a mass of 81kg (ρair=1.3 kg/m3)

The heat energy required to heat the air is: 206.55 kcal over the course of half an hour.

Pair=.1148kcal/sec

Pair=.5kJ/s

Pair=.5kW

2) The rate of heat loss through the walls:

Pwalls=850kcal/hr

Pwalls=1kJ/s

the total power requirements for heating
P = Pair + Pwalls

P = 1.5kW

#### 1 comment:

Desco lab said...