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Physics Department
Year 12 Assessed Homework
Module 4.3.1 Power, series
& parallel
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Total Marks
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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]
RT = 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 = Vtotal/Rtotal
= 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
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
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
= 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
= 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
allow ecf from (iii). (1) 1
