Synoptic question on spectral lines

Questions on Potential dividers markscheme

1.       (i)      M marked at the end of the graph                                                                  B1

(ii)     current is 5 (A) and p.d is 6 (V)                                                                     C1

P = VI \ p = 6.0 ´ 5.0

(Allow p = I2 R or p = V2 \ R)                                                                         C1

power = 30 (W)                                                                                             A1

(iii)    1.      VL = 1.0 (V) (From the I/V graph) \ RL = 1.0/2.0 or 0.5 (W)                  M1

VR = 1.2 ´ 2.0 \ RT = 1.2 + 0.5                                                             M1

V = 1.0 + 2.4 \ V = 1.7 ´ 2.0                                                                 A1

voltmeter reading = 3.4 (V)                                                                  A0

2.      Vr = 4.5 - 3.4 (= 1.1 V) \ 4.5 = 2.0r + 3.4 (Possible ecf)                      C1

r = 0.55 (W)             (1.05 W scores 0/2 since the lamp is ignored)       A1

[9]

 

  2.     (i)      p.d across 1.5 k W resistor = 5.0 - 1.2 = 3.8 (V)                                            B1

(ii)                                   C1

                             C1

R = 474 (W) » 470 (W)                                                                                   A1

(Using 3.8 V instead of 1.2 V gives 4.75 kW - allow 2/3)

[4]

 

3.       (i)      Any four from:

1. The resistance of the thermistor decreases (as temperature is increased)    B1

2. The total resistance (of circuit) decreases                                                   B1

3. The voltmeter reading increases                                                                 B1

          4. Explanation of 3. above in terms of ‘sharing voltage’
/                                                                             B1

5. The current increases / ammeter reading increases                                     B1

6. Explanation of current increase in terms of                                 B1

          (Allow ecf for statements 3. and 5. if statement 1. is incorrect - maximum
score of 2/4)

(ii)                              C1

                       C1

R = 467 (W) » 470 (W)                                                                                   A1

(When 1.4 V and 3.6 V are interchanged, then R = 3.1 ´ 103 (W) can score2/3)

(Calculation of total circuit resistance of 1.67 ´ 103 (W) can score 2/3)

(Use of  scores 0/3)

[7]

 

  4.     (i)      (NTC) thermistor                                                                                            B1

(ii)     Resistance decreases when temperature is increased. (ora)                            B1

(Allow correct credit for a PTC thermistor)

(iii)    1       I = (0.032-0.006 =) 0.026 (A)                                                              B1

2       (V200 = 0.026 ´ 200 =) 5.2 (V) / (V720 = 0.006 ´ 700 =) 4.2 (V)           C1

         E = 5.2 – 4.2                  (Allow E = 4.2 – 5.2)                                      C1

         E = 1.0 (V)                    (Allow 1 sf answer)                                        A1

         (9.4 (V) scores 1/3)

[6]

 

  5.     (i)      The resistance of LDR/circuit changes (as light intensity changes)                 B1
When blade blocks light, resistance of LDR/circuit is large(r) (ora)               B1
Correct statement about p.d (Possible ecf)                                                     B1

(ii)     1.      (V = 5.0 – 3.0)
2.0 (V)        (Allow 1 sf answer)                                                           B1

2.      V =             I = 2.0/2200 / 9.1 ´ 10–4 (A)                           C1
(3.0 = )   (R = 3.0 / 9.1 ´ 10–4)
R = 3300 (W)                 R = 3300 (W)        Possible ecf                        A1
(For VLDR = 2.0 V, R = 1.47 kW. This scores 1/2)
(If 3.5 V given in (ii)1., then R = 940 W. This scores 2/2)

[6]

 

 

current in solids answers

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

 

Markscheme for stellar evolution

1.       (i)      volume = 4π (10,000)3 /3 = 4.2 × 1012 (1)
density = 3.5 × 1030 / 4.2 × 1012 ecf (1)
density = 8.4 × 1017 kg/m3 (1)                                                                          3

  (ii)   any two from
density (very) much greater than material on Earth (1)
quotes typical density on Earth 1 – 104 kg m–3 (1)
atomic structure collapsed / density same as atomic nucleus (1)                       2

[5]

 

  2.     Hydrogen atoms/particles (1)
Collapse under gravity/ decrease of gpe (1)
Increase in kinetic energy/ temperature (1)
Fusion of protons (1)
Energy released/ ref. to E = ∆mc2 (1)

[5]

 

  3.     any 4 from:
end of H burning/red giant/supergiant (1)
onset of He fusion/fusion of heavier nuclei (1)
gravitational collapse of core (1)
supernova explosion/ star explodes (1)
suitable mass limit (chanderasekha limit 1.4M) (1)
supported against gavity by neutron gas pressure/ ref to
Fermi pressure (1)
internal structure protons and electrons combined/ very
thin atmosphere/ metallic crust (1)                                                                             4

[4]

Fuses

Calculate the correct fuse for the appliance below. In all cases voltage = 240  V and fuses available are 3A 5A 13A

 

1.         A 60 W bulb in a table lamp.

2.         A 2.4 kW kettle.

3.         A 1200 W electric fire.

4.         A 480 W television.

5.         A 720 W toaster.

6.         (a) A 100 W microprocessor, a 240 W monitor, a 60 W disc drive, and a  120 W printer.  (b) If all the above were plugged into a multiblock outlet what fuse would you use in that?

Paying for Electricity answers

1.   21.6p            18.1p

2.   £1.16

3.   0.54kWh

4.   .42 kWh        .112p + 3.79p = 3.9p

5.   4.46p            9.34p

Electric Power 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 Power

Power and Energy

1. 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,
2. 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?
3.  Question 1 asks `what is its resistance IN USE', Explain why the last two words have been added.
4.  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)
5.  A current of 2 A flows through a resistor of 10 Ω for 2 min, how much energy is used?
6. 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)
7. 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?
8. 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.
9. 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 !)

Currents in Solids

1. 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?
2. 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.
3. 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?
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 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.