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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