Electric Fields Past Paper Questions

Y12 Electric Power

Power and Energy

A 3 kW immersion heater is designed for use on 240 V mains. Find (a) the current taken from the mains, (b) the resistance of the heater, and (c) the energy used in 1 min,


A 2 kW electric fire is designed for use on 250 V mains. What current does it take, and what is its resistance in use?

Question 1 asks `what is its resistance IN USE', Explain why the last two words have been added.


How much energy is stored in a 12 V car battery of capacity 38 A h ? (ie it is capable of delivering 38A for 1 hour before it is flat)

A current of 2 A flows through a resistor of 10 Ω for 2 min, how much energy is used?


How many lamps marked 175 W 240 V could be used in a circuit fitted with a 5 A fuse if they are connected in parallel? (Remember all components in parallel have the same voltage)

An immersion heater marked 3 kW is designed for use on 240 V. If its resistance remains constant, what is its power if connected to a 200 V supply?


A fire of resistance 30 Ω is connected to a 240 V supply by (a) cables of negligible resistance, (b) cables of total resistance 2 Ω. Find the power consumed by the fire in each case.

A 60 W heater is to be made from manganin wire with a resistance of 3.75 ohms per metre. What length of wire is needed if the heater is to operate on a 12 V supply?

Extra
Find the power wasted in the cables when 120 kW is supplied through cables of resistance 0.2 Ω (a) at 240 V, (b) at 200 000 V.
( Hint answer for (a) is 50 kW, and (b) is much smaller !)


Answers

1 (a)     P = VI               P power in W

                                    V p.d. in V       

                                    I current in A

3000 = 240 x I

I=3000/240 =12.5A

(b)      V = IR

240=12.5R

R = 240/12.5 = 19.2 Ω

(c)      3 kW = 3000 W = 3000 joules per second

Energy used in 1 min = 3000 x 60 = 180 000 J

2        8A       31.25Ω

3        Resistance depends upon temperature

4        1 641 600 J

5        4800J

6        6

7        2.083 kW

8        1920W 1687.5 W

9        .64m

10      50kW  0.072W

Electric Field

Gravitation Field Lines

Gravitation Potential

1. -6.25 x 10^7 Jkg^-1
2.  -6.25 x 10^7 Jkg^-1 : -6.16 x 10^8 J : 1340 ms^-1
3. 1.91 x 10^11 Jkg^-1 ; 6.18 x10^5 ms^-1
4. 9.80 Jm^-1: 1.62 Jm^-1
5. 1.12 x 10^4 ms^-1 ; 2.37 x10^3 ms-1

G and Centrepetal force

Sun Earth

Angular Speed Satellite

Gravitational Field Strength

Electric Current

1. 36C
2. 500s
3. 12A
4. 66 weeks
5. 190mA  27 000C

SHM Revision

Charge and Current

Charge and Current

•The charge carriers are usually electrons; q = e=  -1.60 x 10^-19 C; but be careful when a current of ions exists.

The value of n for copper is 8.0 x 10²⁸ m³.

1. Calculate the current in a copper wire with area of cross section 2.0 x 10^-5 m² when electrons drift through it with a mean speed of 0.80 mms^-1

I = nAve = (8.0 x 10²⁸) (2.0 x 10^-5) (0.80 x 10^-3) (-1.60 x 10^-19) = -205 A

2. In a silicon transistor with area of cross section 3.8 x 10^-6 m² there is a d.c. current of 200 mA. The current is a flow of electrons and the number density of free electrons for silicon is 8.3 x 10²³ m³. Determine the mean drift speed of the electrons.

I = nAve

rearranging v = I/nAe

v = (200 x 10^-3) / (8.3 x 10²³) (3.8 x10^-6) (-1.60 x 10^-19)

v = -0.40 ms^-1

3. The beam of electrons in a cathode-ray tube contains electrons travelling at 8.4 x 10⁶ ms^-1 and the current through the tube is 2.8 μA.
   (a) How many electrons are emitted per second from the cathode of the tube?

Q= It = 2.8 x 10^-6 x 1 second = 2.8 x 10^-6 C

Number of electrons = total charge / e = 2.8 x 10^-6 /1.6x 10^-19  

= 1.75 x 10¹³

(b) What is the number of electrons per unit length of the beam?

In 1 s an electron will travel 8.4 x 10⁶ m. this means that 1.75 x10¹³ will be spread over a length of 8.4 x 10⁶ m. So 1 m will contain 1.75 x10¹³ electrons ÷ 8.4 x 10⁶ m = 2.08 x 10⁶ electrons

4. A direct current of 3.0 A through a copper wire reaches a place where the area of cross section of the wire changes from 2.0 x 10^-6 m² to 0.090 x l0^-6 m². By what factor does the drift speed of the delocalised  electrons increase as they move from the wide to the narrow section? Suggest how this indicates that a damaged wire will overheat.

The same current flows through the thick wire as the thin wire, (Kirchoff’s first law)

For the thick part of the wire

I = nAve

rearranging v = I/nAe

v = 3 / (8.0 x 10²⁸) (2 x10^-6) (-1.60 x 10^-19)

v = -1.17 x 10^-4 ms^-1

For the thin part of the wire

I = nAve

rearranging v = I/nAe

v = 3 / (8.0 x 10²⁸) (0.090 x10^-6) (-1.60 x 10^-19)

v = -2.60 x 10^-3 ms^-1

Ratio = -2.60 x 10^-3 ms^-1 / -1.17 x 10^-4 ms^-1 = 22.2

Alternatively

I = nA_wv_we = nA_nv_ne

As n and v are constants then A_wv_w = A_nv_n

rearranging A_w/A_n=v_n/v_w      A_w = 2.0 x 10^-6 m² and A_n = 0.090 x l0^-6 m²

so A_w/A_n = 2.0 x 10^-6/ 0.090 x l0^-6 = v_n/v_w = 22.2

This increase in drift speed will cause more collisions per second between electrons and the crystal lattice transferring more energy to heat

5. Explain why a light comes on almost immediately when switched on, although the drift speed of the electrons in the wires to the light is so small.

When the circuit is made the electric field propagates through the wire at close to the speed of light. This imposes a net direction from negative to positive on the hitherto random motion of the delocalised electrons

Currents in solids

Some copper fuse wire has a diameter of 0.22mm and is designed to carry currents of up to 5.0 A. If there are 1.0 x 10²⁹ electrons per m³ of copper, what is the mean drift speed of the electrons in the fuse wire when it carries a current of 5.0A?

I = nAve rearranging v = I/nAe             Area = ∏r² r = 0.22 x 10^-3 /2 so ∏r² = ∏(0.11 x 10^-3)² = 3.8 x 10^-8 m²

So v = 5.0/ (1.0 x 10²⁹)( 3.8 x 10^-8)(1.6 x 10^-19) = 8.2 x 10^-3 ms^-1

A wire carrying an electric current will overheat if there is too large a current: the accepted value for the maximum allowable current in a copper wire is 1.2 x 10⁷ A per square metre of cross-section of the wire. If there are 1.0 x 10²⁹ electrons per m³ of copper, calculate the mean drift speed of the electrons in the wire when the current reaches this value.

I = nAve rearranging v = I/nAe            

So v = 1.2 x 10⁷ / (1.0 x 10²⁹)(1)(1.6 x 10^-19) =  7.5 x 10 ^-4 ms^-1 = 0.75 mms^-1

Two copper wires of diameter 2.00 mm and 1.00mm are joined end-to-end. What is the ratio of the average drift speeds of the electrons in the two wires when a steady current passes through them? In which wire are the electrons moving faster?

I = nAve rearranging v = I/nAe             Area = ∏r² r

For d = 1.00 mm r = 1 x 10^-3 /2 so ∏r² = ∏(0.5 x 10^-3)² = 7.85 x 10^-7 m²

I don’t know the current so I, but I do know it is the same in both so let’s say it is 1A

So v = 1/ (1.0 x 10²⁹)( 7.85 x 10^-7)(1.6 x 10^-19) = 7.9  x 10^-5ms^-1

For d = 2.00 mm r = 2 x 10^-3 /2 so ∏r² = ∏(1 x 10^-3)² = 3.14 x 10^-6 m²

I don’t know the current so I, but I do know it is the same in both so let’s say it is 1A

So v = 1/ (1.0 x 10²⁹ (3.14 x 10^-6)(1.6 x 10^-19) = 1.99  x 10^-5ms^-1

Ratio of 2mm/1mm = 1.99/7.9 = 0.25 = 1:4

Alternate

I = n₁A₁v₁e₁ = n₂A₂v₂e₂           As e and n are constant A₁v₁ = A₂v₂ and v₂/v₁ = A₁/A₂ 

as A = ∏r² = ∏(d/2)²               then A₁ = ∏(1/2)²  and A₂ =∏(2/2)² so A₁/A₂ = ∏(1/2)²/∏(2/2)² = 1/4

A copper wire joins a car battery to one of the tail lamps and carries a current of 1.8A. The wire has a cross-sectional area of 1.0 mm² (1x 10^-6m²)and is 6.0 m long. If there are 1.0 X 10²⁹ electrons per m³ of copper, calculate how long it takes an electron to travel along this length of wire.

I = nAve rearranging v = I/nAe            

So v = 1.8 / (1.0 x 10²⁹)(1 x 10^-6)(1.6 x 10^-19) = 8.2 x 10^-3 ms^-1 = 1.125 x 10 ^-4 ms^-1

 v = s/t rearranging t = s/v = 6m / 1.125 x 10 ^-4 ms^-1 = 5.33 x 10⁴ s = 14.8 hrs

SHM Experiment with a Spring