Capacitors EUT markscheme

A* - 31

A – 29

B – 25

C – 22

D – 18

E – 14

 

1.       (i)      Cp = C + C = 6 μF; 1/Cs = 1/2C + 1/C; = 3/2C giving Cs = 2C/3 = (2 μF)          3

(ii)     2 sets of (3 in series) in parallel/ 3 sets of (2 in parallel) in series                     2

[5]

 

  2.     (a)     (i)      Cp = 2 + 4 = 6 μF                                                                                  A1

  (ii)   1/C = 1/2 + ¼                                                                                        C1
Cs = 4/3 =1.33 μF                                                                                 A1

  (b)   (i)      6.0 V                                                                                                    A1

  (ii)   Q = CpV                                                                                                C1
= 6 × 6 = 36 μC                                                                                    A1

  (c)   E = ½ CsV2                                                                                                    C1
= 24 × 10–6                                                                                                    A1

  (d)   (i)      The capacitors discharge through the voltmeter.                                   B1

  (ii)   V = V0e–t/CR
1/4 =e–t/(6×12)                                                                                        C1
ln 4 = t / 72                                                                                           C1
t = 72 ln 4 ≈ 100 s                                                                                A1

[12]

 

  3.     (i)      C = Q/V or gradient of graph / = 24 μC/3V; = 8.0 (μF)                                     2

(ii)     E = ½ CV2 / = ½ × 8 × 32; = 36 (μJ) ecf a(i)                                                      2
or ½ QV / = ½ × 24 × 3; = 36 (μJ)

  (iii)  T = RC = (0.04); R = 0.04/8.0μ = 5.0 × 103 (Ω) ecf a(i)                                    2

(iv)    idea of exponential/constant ratio in equal times; which is independent of
initial value/AW or argued mathematically in terms of Q/Qo = e–t/RC
give 1 mark for statement that time depends only on time constant/RC            2

[8]

 

  4.     (a)     (i)      Q = VC; W = ½ VC.V ( = ½ CV2) (2)

(ii)     parabolic shape passing through origin (1)
plotted accurately as W = 1.1 V2 (1)                                                       4

  (b)   (i)      T = RC; = 6.8 × 103 × 2.2 = 1.5 × 104 s = 4.16 h                                     2

(ii)     ΔW = ½ C(V12 –V22) = 1.1(25 – 16) ; = 9.9 (J)                                       2

  (iii)  4 = 5 exp(–t/1.5 × 104) ; giving t = 1.5 × 104 × ln 1.25 = 3.3 × 103 (s)    2

(iv)    P = ΔW/Δt = 9.9/3.3 × 103 = 3.0 mW            ecf b(ii) and (iii)                 1
allow P = Vav2 /R = 4.52/6.8 × 103 = 2.98 mW

[11]

 

 

Assessed homework materials

26 A

23 B

20 C

17 D

13 E

1.       (i)      Stress = force / cross-sectional area                                                                B1

(ii)     Strain = extension / original length                                                                 B1

[2]

2.       (a)     Young modulus = stress/strain
(As long as elastic limit is not exceeded)                                                        B1

(b)     Strain has no units because it is the ratio of two lengths.                                B1

[2]

3.       (a)     The extension of a spring is directly proportional to the applied force          M1
as long as the elastic limit is not exceeded)                                                    A1

  (b)   (i)      Correct pair of values read from the graph
force constant = 12/0.080                                                                     C1
force constant = 150 (N m–1)                                                               A1

(ii)     extension, x =  × 80 (= 133.33) (mm)                                              C1
(E = ½ Fx)
energy = ½ × 2/ × 133.33 × 10–3
energy = 1.33 (J)                                                                                  A1

(iii)    The spring has not exceeded its elastic limit                                         B1

(iv)    (elastic potential energy = kinetic energy)
                                                                                      M1
m and k are constant, therefore x µ v.                                                  M1

[9]

4.       (a)     A brittle material does not have a plastic region / it breaks at its elastic limit. B1

(b)     Ultimate tensile strength is breaking stress for a material                               B1
Materials can be chosen / tested to prevent collapse of the bridge                 B1

[3]

5.       (i)      1       Elastic as returns to original length (when load is removed)                  B1

2        Hooke’s law is obeyed as force is proportional to the extension          B1

Example of values given in support from table                                    B1

(ii)     Measure (original) length with a (metre) rule / tape                                        B1

Suitable method for measuring the extension e.g.

levelling micrometer and comparison wire or fixed

scale plus vernier or travelling microscope and marker / pointer                    B1

(iii)    E = stress / strain                                                                                            C1

    = (25 ´ 1.72) / (1.8 ´ 10–7 ´ 1.20 ´ 10–3)                                                   C1

    = 1.99 ´ 1011 (Pa)                                                                                       A1

[8]

6.       (a)     P.E. at top = 80 × 9.8(1) × 150 = 118 000 (J) (1)
K.E. at bottom and at top = 0 (1)
Elastic P.E. at top = 0, at bottom = P.E. at top for ecf = 118 000 J (1)              3

  (b)   24 N m–1 × 100 m = 2400 N                                                                            1

  (c)   elastic P.E. is area under F-x graph (1)
graph is a straight line so energy is area of triangle (1)
elastic P.E. = ½ × kx × x = (½kx2) (1)                                                               2

  (d)   loss of P.E. = 100 × 9.8(1) × 150 = 147 000 J (1)
gain of elastic P.E. = ½ × 26.7 × 1052 = 147 000 J (1)                                      2

  (e)   idea that a given (unit) extension for a shorter rope requires a greater force     1

[9]