y12 Assessed homework Waves

y12 Assessed Hw

Physics Department Year 12 Assessed Homework Module 4.3.1 Power, series & parallel

Total Marks

In class assessed questions

1. The diagram shows a resistor network.

The total resistance between points X and Y is [1] C

    A  0.25 Ω

    B  1.0 Ω

    C  4.0 Ω

    D  16 Ω

2. Two identical resistors connected in series have a total resistance of 8 Ω. The same two resistors when connected in parallel have a total resistance of [1] B

    A  0.5 Ω

    B  2 Ω

    C  4 Ω

    D  8 Ω

3. Define the term power [1] rate of transfer of energy

4. Write down three equations you could use to calculate electrical power [3] P=IV [1]  P=I²R  {1]  P=V²/R [1]

5. State Kirchoff’s laws [2] K1 – Total current entering a junction = total current leaving it [1]

K2 – total emf around a series circuit = the sum of the p.d.s across each component [1]

3. How are resistors R₁ and R₂ connected if their total resistance is equal to R₁+R₂? [1]

Series [1]

4. A battery provides 3400J of energy per second. What is the power of the battery? [1] 3400W

Teacher assessed questions

1. A resistor is connected to a cell. An amount of charge Q passes through the resistor in a time t. During this time, the amount of chemical energy converted to electrical energy by the cell is E. Select the row of the table which correctly gives the current in the resistor and the e.m.f. of the cell. [1] C

2. Which of the following can be used as a unit of electrical resistance? [1] A

    A  W A⁻²

    B  A V⁻¹

    C  W V⁻²

    D  V C⁻¹

3. Which combination of resistors has the smallest total resistance? [1] D

    A  

    B  

    C  

    D  

4. A car starter motor requires 12.5kJ of energy to flow through it in 2.00 seconds to start the engine.

                a) Calculate the power necessary to start the engine [1]

P = W/t = 12500/2.00 = 6250W

                b) The car battery supplies 8.00V to the starter motor. Calculate the current required to start the engine [1]

I=P/V = 5200/230 = 23A

5. A circuit in an electric car converts 1250J of electrical energy into heat every second. The resistance in that circuit of the car is 54.2W. Calculate the current through that circuit. [2]

 [1]

I = 4.80A [1]

6. The battery in the circuit below has a negligible internal resistance. The total resistance in the circuit is 10W.

                a) Calculate the resistance of R₁. [3]

R (3 & 6 W)

1/R = 1/3 + 1/6 = 1/2  therefore R = 2W [1]

R_T = 10W  10 = uknown + 5 + 2, so unknown = 3W [1]

R (12 & R₁)

1/3=1/R₁ + ¼ à R = 12W [1]

                b) Calculate the potential difference across the 5W resistor [2]

I = V_total/R_total = 12/10 = 1.2A [1]

V = I x R = 1.2 x 5 = 6V [1]

                c) Calculate the current through the 6W resistor. [2]

p.d. across the resistors in parallel (3 & 6) à V = 1.2 x 2 = 2.4V [1]

so current through 6W is à I=V/R = 2.4/6 = 0.4A [1]

Past exam questions [9]

1. The figure below shows a network of identical resistors.

Calculate the total resistance between points A and B. [3]

total resistance of three in series = 6.0 (kW)                                                                          

              \                                                                                         

resistance = 1.5 (kW)                                                                                                   

2. This question is about the design and use of Christmas tree lights.

          Design of bulbs

          An engineer intends to design light bulbs for use in a set of Christmas tree lights to be powered by a 240 V mains supply. Each bulb, when operating normally, will use 0.50 W and will have a filament 6.0 mm long, made of tungsten. The resistivity of tungsten at normal working temperature = 1.1 × 10–6 Ω m

(a)     State one advantage of connecting these bulbs in parallel, rather than in series. [1]

(b)     Suppose the bulbs are connected in parallel. Calculate

(i)      the current through each bulb  [2]

(ii)      the resistance of each bulb filament [2]

(iii)     the radius of each bulb filament. [3]

(iv)    Hence suggest why these bulbs are impractical. [1]

(a)     either (If in parallel) when one bulb fails, other bulbs stay on
or         (If in parallel) can identify which bulb has failed; (1)                                     1

  (b)   (i)      P = VI (1)
0.5 = 240 I
I = 2.1 × 10–3 A 1 s.f. in answer (–1) once only (1)                                 2

  (ii)    R = V/I (1)
= 240/(2.1 × 10–3)
= 1.14 × 105 Ω or 1.15 × 105 Ω ans
accept (1.1 to 1.2) × 105 Ω. (1)                                                                        2

  (iii)   A = ρ l / R (1)
= 1.1 × 10–6 × 6.0 × 10–3 / (1.14 × 105) (= 5.79 × 10–14 m2)
A = πr2 (1)
5.79 × 10–14 = πr2 so r = 1.4 × 10–7 m (1)                                                       3

  (iv)  filament too thin / fragile to be manufactured / used without damage;
allow ecf from (iii). (1)                                                                                      1

Induced current in a Loop of wire

y13 Flux Induction Question

y12 Intensity

Magnetic Field in a Wire

Nitrogen ion in an electric field

(a)     Positive as E-field is downwards/top plate is positive/like charges repel/AW (1)       1

  (b)   (i)      k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2)                                2

  (ii)    1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1) (2)                                                                2

F= BIL

Emf and internal resistance 2 (y12)

y12 Thermistor 3

Electric Field Exam Questions (all of them in no particular order)

Alpha Particle Charged plates

E Field calc

E Field Bunsen Flame

Electron in E Field 3

Electron in E Field

Electron in E Field 2

Experiment with scales and rods

Force in Field

Internal Resistance of a cell

Internal Resistance

Q 1,2 & 3 Look up the answers in your text books.

Q4       Total R = 2 + 5.5

            I =  1.5 V/Total R

            Pd across cell = pd across external R = I(5.5) = 1.1V

Q5       emf = 1.5V

            Lost volts = (1.5 – 1.2) V

Internal r = lost volts/0.30A = 1.0 ohm

External R = 1.2V/0.30A = 4.0 ohm

Q6       lost volts = 100A (0.04 ohm) = 4V

            Only 8V across external circuit so bulbs are dim

Q7       I=V/R= 1.25V/25 ohm = 0.05A

            Internal r = lost volts/I = 0.25V/0.05A = 5 ohms

            Total R with 10 ohm = 10 + 5 =15

            I = E/R = 1.5/15 = 0.1A

            V across terminals= V across 10 ohm R = 0.1A(10 ohm) = 1V

Coulomb's Law Practical

Electric Power and Cost

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.     21.6p  19.1p

11      £1.16

12      0.54kWh

13      .42 kWh        .112p + 3.79p = 3.9p

14      4.46p            9.34p

Extra  50kW  0.072W

Nitrogen Atom

1.       (a)     Positive as E-field is downwards/top plate is positive/like charges repel/AW (1)  1

  (b)   (i)      k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2)                                2

  (ii)   1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1) (2)                                                             2

  

Ke of Satellite

Slight mistake here left an m out of the 1/2 GMm/r

oops - forgot to multiply by the 10kg for Ke so that should be 2.01 x 10^8 J
 and GPE is for 1km = 1000m so GPE = 0.402 x1000 = 402J

Gravity Apollo parking orbit

Gravity QUestions

 1.      (i)      r has been increased by a factor of 3 from the centre of planet.                    C1
g = (40/32 =) 4.4(4) (N kg–1)                                                                          A1

  (ii)   M = gr2 / G
M = (40 × [2.0 × 107]2) / 6.67 × 10–11                                                            C1
M = 2.4 × 1026 (kg)                                                                                        A1

  (iii)  M = ρV = 4/3 πr3 ρ                                                                                         M1
g = GM / r2  r3 / r2 (Hence g  r)                                                              A1

[6]

2        The astronaut is accelerating / has centripetal acceleration (1)
and the space station has the same acceleration (1)
a person does not feel gravity (1)
only feels forces applied by contact with the walls of the space station (1)
no support force from the space station (as they have the same acceleration) (1)      4

MAXIMUM (4)

[4]

 3.      Period 24 hours (1)
Satellite must stay locked into Earth’s period of rotation (or wtte) (1)

          Plane Equatorial (1)
Centre of orbit must be centre of Earth because axis of orbit must be spin axis of Earth (1)
(gravitational force above equator is only force available to provide
centripetal force in a synchronised orbit, otherwise an engine is required)

          Direction Same as Earth’s rotation (1)
(otherwise satellite and Earth would counter rotate)

[5]

  4.     (a)     i. F = GMm/r2 or F α Mm/r2 with labels (1)                                                       1
ii. finite universe contracts/ resultant force on stars (1)                                    1

  (b)   Any 2 from
i. (satellite B) has larger circumference/smaller velocity
(satellite B) Gravitational field strength is less
(satellite B) Centripetal force is less                                                                  2

          ii.(accept calculation from either satellite)
r13/ T12 = r23/ T22 (1)
satellite A                                satellite B
r23 = 70003 × 57.22 / 1.632      r23 = 671003 × 57.22 / 1.632 (1)
r2 = 75,030 km                        r2 = 75, 320 km (1)                                           3
(= 75,000 km)                         (= 75,000 km)

  (c)   Land-based are (any 3) 1 mark for each
more light can be collected/ made larger
more stable
more manoeuvrable
cheaper to build/repair
longer lifetime/ not exposed to high velocity particles
greater access                                                                                                   3

[10]

  8.     (i)      4.5 (N kg–1)                                                                                                      1

(ii)     g = (–)GM/r2                                                                                                     1

(iii)    g ∞ 1/r2;so value is 40/9 = 4.4(4) (N kg–1) ecf c(i)                                            2

[4]

× 2.0 × 107/(4.5 × 103) = 2.8 × 104 m s–1 (2)
½ mv2; = 0.5 × 1500 × (2.8 × 104)2 = 5.9 × 1011 (J) (2)                                             4
aliter: F = mv2/R; = mg; so ½ mv2 = ½ mgR;= 6.0 × 1011 (J)

[8]

y12 Nov Ass

Part A

1. D
2. E
3. C
4. A
5. D
6. A
7. A
8. B
9. D

Part B

SHM

Divided Circuits

1. Two resistors of 3Ω  1.4A and 7Ω 0.6A are connected in parallel. If the total current through them is 2 A find the current in each resistor. 

2. Two resistors of 2 Ω 2A and 4 Ω 1A are connected in parallel. If the current through them is 3 A find the current in each resistor.

3. Three resistors of 2 Ω 4 Ω 0.5A and 8 Ω 0.25A are connected in parallel in a circuit. The current in the 2 Ω resistor is 1 A. What is the current in the other two resistors? What is the total current in the circuit? 1.75A

4. Three resistors of 4Ω 0.5A, 5Ω 0.4A and 20Ω 0.1A are connected in parallel. If the total current through them is 1 A find the current through each resistor.

5. Ammeter X reads 2 A. What are the readings of am­meters A, 4A B1A, C 1A and D 4A?

6. A resistor of 2 Ω and another of 7Ω are connected in series, and a 3Ω resistor is connected in parallel across the other pair. If the total current through the network is 2 A, find the p,d. across the 7Ω resistor. 3.5V

The Volt

Volt

                                V = E/Q                 1 Volt = 1 Joule / 1 Coulomb

1.       Calculate the potential difference across a bulb if 20 C transfers 4 J of electrical energy.

V = 4 J /20 C = 0.2 V

2.       Calculate the emf across a dynamo if 600 J of energy is transferred by 25 C of charge.

V = 600 J / 25 C = 24 V

3.       A bulb has a potential difference of 6V across it. Calculate the energy is transferred by 90 C.

E = V x Q = 6 V x 90 C = 540 J

4.       A train motor operates on a potential difference of 25 kV .Calculate the charge that transfers  800 kJ of electrical energy.

Q = E / V = 800000 J / 25000 V = 32 C

5.       A torch bulb runs off a cell of emf 1.5V Calculate the energy transferred by a charge of 0.27 C

E = V x Q = 1.5 V x 0.27 C = 0.41 J

A bulb takes a current of 0.25 A. How many Coulombs per second? 0.25 C per second It transfers 60W of powerHow many Joules per second? . 60 J per second

Calculate the emf of the supply. V = E ÷ Q = 60 / 0.25 = 240 V

Circular Motion _London Eye

(a)     (i)      speed v = 2π r / t
v = 2 × π × 122/2 /(30 × 60) (1)
v = 0.21 m s–1 (1) allow 0.2 m s–1                                                           2

  (ii)   F = 12.5 kN × 16 = 200 kN (1)                                                                1

(iii)    W = F × s or
= 200 k × 2 × π × 122 / 2 (1) ecf (ii) allow ecf for distance from (i)
= 7.7 × 107 J (1) allow 8 × 107                                                                2

  (iv)  P = W / t, energy / time or F × v or
= 7.67 × 107 / (30 × 60) (1) or ecf (iii) / (30 × 60)
= 42.6 kW (1) allow 43 kW, only allow 40 kW if working shown             2

(v)     •        Friction force at bearing opposes motion so not useful (1)
•        Friction force of tyres on rim drives wheel, so is useful (1)
•        Electrical energy supplies power to drive wheels /
          useful implied (1)
•        Input energy (electrical or energy supplied to motor)
          is converted into heat (1)

          Last point to do with the idea that once moving with constant speed e.g.
•        All work is done against friction
•        No input energy is converted into Ek
•        All input energy ends up as heat
•        Any other relevant point relating to energy (1)                              5

  (b)   (i)      k = F / x
= 1.8 × 106 / 0.90 (1)
= 2.0 × 106 Nm–1 (1)                                                                               2

the pendulum bob is travelling in a circle (1)
so it is accelerating towards the centre (1)
(it has a constant speed in the time interval just before vertical to just
after vertical)

          bob is not in equilibrium (1)
so the tension must be (slightly) larger than the weight of the bob (1)                        3

MAXIMUM 3