16PS. Capacitors (2026) tutorial solutions NJC

National Junior College Science Department | Physics 1 16. Capacitors Problem set (suggested solutions) 1 (a) apply potential difference between the plates M1 causes charge separation between the plates A1 (b) (i) straight line starting at the origin B1 line with positive gradient ending at (V, Q) B1 (ii) work done is the area under the graph B1 W = ½ QV B1 (c) (i) Electric field is constant / remains the same B1 E = VL = VNL!D M1 VN="L−DL$V (ii) Charge on each plate remains the same. B1 Charge on each plate Q = CV C1 CN= QVN = CV"L!DL#V = !LL!D" C A1 2 (a) Combined capacitance = 24 + (1/24 + 1/24 + 1/24)–1 = 24 + 8 = 32 µF (b) (i) parallel combination of two in series and a single capacitor (ii) one capacitor in series with two in parallel

National Junior College Science Department | Physics 2 3 (a) relationship between charges relationship between p.d.s series QS = Q1 = Q2 VS = V1 + V2 parallel QS = Q1 + Q2 VS = V1 = V2 (b) (i) E = ½ CV2 19 × 10–3 = ½ (470 × 10–6)V2 C1 V = 9.0 V A1 (ii) E = Q2 / 2C (or use C = Q / V) 19 × 10–3 = Q2 / (2 × 470 × 10–6) C1 Q = 4.2 × 10–3 C A1 (iii) 1. total capacitance = (470 + 180) × 10–6 = 650 × 10–6 C1 total charge unchanged C1 p.d. across capacitor = p.d. across total capacitance = Q / C = (4.226 × 10–3) / (650 × 10–6) = 6.5 V A1 2. Total energy in the two capacitors = Q2 / 2C = (4.226 × 10–3)2 / (2 × 650 × 10–6) = 13.74 mJ C1 Decrease in total energy = 19 – 13.74 = 5.3 mJ A1

National Junior College Science Department | Physics 3 4 combined capacitance = (1 / 22 + 1 / 47)–1 = 14.99 µF C1 initial charge on 22 µF = initial charge on combined capacitors = 14.99 × 12 = 179.88 µC AND charge on 22 µF when p.d. is 6 V = 22 × 6 = 132 µC C1 132 = 179.88 exp[–t / (2.7 × 106 × 14.99 × 10–6)] C1 Note: For the time constant, it is incorrect to use 22 𝜇F even though we compute the charge using it. This is because the circuit comprises both capacitors and the time constant is due to both capacitors. t = 12.5 s or 13 s A1 5 (a) current in resistor causes capacitor to lose charge, since p.d. proportional to charge so p.d. across capacitor decreases B1 rate of change of p.d. across capacitor decreases as p.d. (or charge) decreases B1 (since !!"(𝑉#𝑒$!"+=−%&𝑉#𝑒$!", gradient of p.d. decreases) since p.d. across resistor = p.d. across capacitor B1 so rate of change of p.d. across resistor decreases as p.d. (or charge) decreases (b) [Method 1] Q0 = 0.90 mC B1 At t = one time constant, Q = 0.90 e–1 = 0.33 mC M1 Construction on Fig. (b) showing when Q = 0.33 mC, 𝜏 = 5.5 s A1 [Method 2] Evidence of two correct sets of readings for Q and t from the graph B1 Correct substitution of Q1 = Q0 exp (–t1 / 𝜏) and Q2 = Q0 exp (–t2 / 𝜏) M1 Correct calculation Q2 / Q1 = exp [(t1 – t2) / 𝜏] to give 𝜏 = 5.5 s A1 [Method 3] Read-off of half-life as 3.75 s B1 Use Q = Q0 exp (–t / 𝜏) to show that 𝜏 = half-life / ln 2 M1 𝜏 = 3.75 / ln 2 = 5.4 s A1 (c) (i) at t = 0, charge on capacitor = 0.90 mC and p.d. = 7.5 V C1 C = 0.90 × 10–3 / 7.5 = 120 µF A1 (ii) R = 𝜏 / C = 5.5 / 120 × 10–6 C1 R = 46 kΩ A1

National Junior College Science Department | Physics 4 6 (a) VC equal VR, so both figures show that Q is proportional to I M1 (VC= Q / C and VR=𝐼R, so Q / C =𝐼R) current is rate of change of charge M1 (𝐼= dQ / dt) so, Q is proportional to rate of change of Q A1 (Q / C = R dQ / dt) this lead to exponential variation of Q (b) from Fig. (b), C = 7.2 × 10–3 / 12 = 0.0006 F C1 from Fig. (c), R = 12 / 1.5 × 10–3 = 8000 Ω C1 time constant = RC = 8000 × 0.0006 = 4.8 s A1

National Junior College Science Department | Physics 5 7 (a) smoothing of output voltage B1 (b) U = ½ CV2 0.041 = ½ C (12)2 M1 C = 569.4 = 570 µF A0 (c) During discharging, p.d. decreases from 12 V to 8 V in 0.01 s C1 V = V0 exp (–t / RC) 8 = 12 exp [–0.01 / (R × 570 × 10–6] M1 R = 43.27 = 43 Ω A1

National Junior College Science Department | Physics 6 Challenging questions C1 Rearranging the capacitors systematically, 𝐶=2+3+6=11 𝜇F C2 The interleaved sheets can be considered as 7 capacitors in parallel. Capacitance of each capacitor 𝐶'=𝜀#(/*!=%*(𝜀#(!+=%*𝐶 𝐶+,,=7×14𝐶=74𝐶

National Junior College Science Department | Physics 7 C3 (a) Total charge 𝑄=2𝐶𝑉# After switch is opened, the capacitors are in parallel. Effective capacitance after adjustment 𝐶+,,=𝐶+𝑓𝐶=𝐶(1+𝑓) Steady state p.d. for both capacitors 𝑉=𝑄𝐶=2𝐶𝑉#𝐶(1+𝑓)=2𝑉#1+𝑓 (b) Before adjustment, energy 𝑈-=%./#0=%.(2𝐶𝑉#).%.0=𝐶𝑉#. After adjustment 𝑈1=12𝐶"2𝑉#1+𝑓$.+12𝑓𝐶"2𝑉#1+𝑓$.=12𝐶"2𝑉#1+𝑓$.(1+𝑓)=2𝐶𝑉#.1+𝑓 Difference in energy Δ𝑈=𝑈1−𝑈-=2𝐶𝑉#.1+𝑓−𝐶𝑉#.=2𝐶𝑉#.−𝐶𝑉#.(1+𝑓)1+𝑓=𝐶𝑉#.(1−𝑓)1+𝑓 Since f < 1, Δ𝑈>0. Final energy > initial energy. Work is done to change the variable capacitor value that increases the energy of the system. C4 (a) 𝐸=%.𝐶%𝑉#. (b) (i) At steady state, p.d. across both capacitors is V and total charge in the capacitors is the same as the charge in C1 before switch S2 is closed. 𝐶%𝑉+𝐶.𝑉=𝐶%𝑉#⇒𝑉=0$0$20#𝑉# Total energy in capacitors at steady state =%.𝐶%𝑉.+%.𝐶.𝑉. (=%.0$#0$20#𝑉#.+. Energy dissipated 𝐸3455=%.𝐶%𝑉#.−%.0$#0$20#𝑉#. ∴𝐸3455=0$0#.(0$20#)𝑉#. (ii) Energy dissipated is total energy of two capacitors at steady state after S2 is closed, so 𝐸′3455=%.0$#0$20#𝑉#.

National Junior College Science Department | Physics 8 C5 (a) (i) p.d. across network of capacitors is 24 V. 𝐶89:="1𝐶+13𝐶$$%=34𝐶 𝑄89:=34𝐶×24=18𝐶 𝑉89=𝑄89:𝐶=18 V 𝑉9:=𝑄89:3𝐶=6 V 𝐶8;:="12𝐶+14𝐶$$%=43𝐶 𝑄8;:=43𝐶×24=32𝐶 𝑉8;=𝑄8;:2𝐶=16 V 𝑉;:=𝑄8;:4𝐶=8 V (ii) 𝐶+<=(%0+%=0+$%+(%.0+%*0+$%=.>%.𝐶 Therefore, the equivalent capacitance 𝐶+< between points W and Y of the network is just over 2C. (iii) 𝐼=𝑑𝑄/𝑑𝑡 𝐼=𝑄#𝑅𝐶+<𝑒$ "@0%&=𝐸𝑅𝑒$ "@0%&=𝐼#𝑒$ "@0%& 𝐶+<=2512𝐶=2512"5004$=312512 𝜇F 12𝐼#=𝐼#𝑒$ A@0%& ln12=−𝑇𝑅𝐶+< 𝑇=𝑅𝐶+<ln2=(130×10=)"312512×10$B$ln2=23.5 s (iv) 𝐶=>##*=125 𝜇F 𝑄=𝐶𝑉⇒𝑞=(125×10$B)(18)=2.25×10$= C (v) Energy stored =%.𝐶+<𝑉.=%.(=%.>%.×10$B+(24).=0.075 J

National Junior College Science Department | Physics 9 (b) Notice the group of 3 capacitors joining to point A and another group of 3 capacitors joining to point G. The other 6 capacitors joins the two groups. The network of capacitors can be redrawn as (i) BDE or CFH (ii) 𝐶C=(%=×.>+%B×.>+%=×.>+$%=30 𝜇F C6 (a) max charge stored in one cycle Q = CV and period T = 1/f C1 substitution into I = Q / t, 2.5 × 10–6 = C × 180 / (1 / 50) M1 C = 280 pF A1 (b) total / equivalent capacitance increases M1 greater charge for each cycle / discharge so greater average current A1 A (B, D, E) (C, F, H) G

National Junior College Science Department | Physics 10 C7 (a) P labelled “–” and Q labelled “+” B1 (b) VOUT scale labelled 4 and 8 at the tick marks B1 T = 2𝜋 / 𝜔 = 2𝜋 / 25𝜋 = 0.08 s B1 t scale labelled 0.02, 0.04, 0.06, 0.08, 0.10, 0.12 on the tick marks B1 (c) (i) Correct symbol used for capacitor and capacitor connected in parallel with the 1.2 kΩ resistor. B1 (ii) Straight lines (or curve with negative decreasing gradients) drawn between adjacent peaks, from top of the first peak to meet line going up to the next peak. B1 Line, from one peak to the line going up to the next peak, show a drop in p.d. of 1.5 small squares. B1 (iii) minimum VOUT = 0.90 × 6 = 5.4 V AND discharge time = 0.034 s C1 V = V0 exp (–t / RC) 5.4 = 6 exp [–0.034 / (1.2 × 103 × C] M1 C = 2.7 × 10–4 F A1