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Monday, November 08, 2021

Past Questions on Capacitance

 

1.       (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 = V0et/CR
1/4 =et/(6×12)                                                                                          C1
ln 4 = t / 72                                                                                            C1
t = 72 ln 4 ≈ 100 s                                                                                  A1

[12]

 

  2.     (a)     Qo = CV = 1.2 × 10–11 × 5.0 × 103; = 6.0 × 10–8; C (3)                                         3

  (b)   (i)      RC = 1.2 × 1015 × 1.2 × 10–11 or = 1.44 × 104 (s) (1)                                   1

(ii)     I = V/R = 5000/1.2 × 1015 or = 4.16 × 10–12 (A) (1)                                    1

  (iii)  t = Qo/I; = 6 × 10–8 / 4.16 × 10–12 = 1.44 × 104 (s)                                       2

  (iv)  Q = Qoe–1; Q = 0.37Qo so Q lost = 0.63Qo                                                  2

  (c)    (i)      capacitors in parallel come to same voltage (1)
so Q stored α C of capacitor (1)
capacitors in ratio 103 so only 10–3 Qo left on football (1)                            3

  (ii)   V = Q/C = 6.0 × 10–8 /1.2 × 10–8 or 6.0 × 10–11 /1.2 × 10–11 or only 10–3
Q left so 10–3 V left; = 5.0 (V)                                                                   2

[14]

 

  3.     (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]

 

  4.     (a)     29; 34                                                                                                                2


(b)     λ = 0.693/T = 0.693/(120 × 3.2 × 107) = (1.8 × 10–10 s–1) accept ln 2                     1

  (c)    (i)      Q = CV = 1.2 × 10–12 × 90; evidence of calculation (= 1.1 × 10–10 C)           2

  (ii)   n = Q/e = 1.1 × 10–10/1.6 × 10–19; = 6.9 × 108 allow sig. fig. variations        2

(iii)    A = λN; N = 6.9 × 108/1.8 × 10–10; = 3.8 × 1018 using 7.0 gives 3.9             3

  (iv)  1 y is less than 1% of 120 y so expect to be within 1%/
using e–λt gives exactly 1% fall/ problem of random emission
or other relevant statement                                                                        1

[11]

 

  5.     (a)     (i)

capacitor

capacitance / µF

charge / µC

p.d. / V

energy / µJ

X

5

30

= Q/C

= 6 (V) (1)

= ½ CV 2(1)

= ½ × 5 × 62

= 90 (1)

Y

25

= CV

= 25 × 6

=150 (µC) (1)

= 6 (V) (1)

= 450 (1)

Z

10

30 + 150 =

180 (µC) (1)

= Q/C

= 180/10

= 18 (V) (1)

= 1620 (1)

         Each box correctly calculated scores (1) + (1) for ½ CV2                             9

  (ii)   1   18 V + 6 V = 24 (V) (1)

2   180 (µC) (1)

3   180 / 24 = 7.5 (1)

4   90 + 450 + 1620 = 2160 (µJ) (1)                                                           4

  (b)   (i)      Kirchhoff’s second law OR conservation of energy (1)                               1

(ii)     Kirchhoff’s first law OR conservation of charge (1)                                    1

  (c)    (i)      time constant = CR (1)
= 7.5 × 10–6 × 200 000 = 1.5 (s) (1)                                                            2

  (ii)    (1)
Q/Qo = e–4 = 0.0183 (1)                                                                             2

[19]

 

  6.     (a)     (i)      5.0 (V) (1)                                                                                                1

(ii)     10.0 (V) (1)                                                                                              1


(b)     (i)      Q = CV;= 1.0 × 10–3 (C) (2)                                                                      2

(ii)     The total capacitance of each circuit is the same (namely 100 μF); (1)
because capacitors in series add as reciprocals/ in parallel add/
supply voltage is the same and Q = VC, etc. (1) max 2 marks                      2

  (c)    (i)      A1 will give the same reading as A2; because the two ammeters are (1)
connected in series /AW (1)                                                                      2
answer only in terms of exponential decrease for a maximum of 1 mark

  (ii)   A4 will show the same reading as A2 at all times; (1)
A3 will show half the reading of A2 initially;
and at all subsequent times (2)                                                                   3

[11]

 

  7.     (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]

 

  8.     (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]

 

 

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