16ES. Capacitors (2026) notes NJC exercise solutions

National Junior College Science Department | Physics 1 Topic 16: Capacitors Solutions to Exercises Exercise 1 1.1 An uncharged capacitor is connected between earth Z and a terminal W. A positively charged rod is brought close to W. Which of the following describes the movement of charge? A Electrons move from W to X and from Y to Z. B Electrons move from W to X but not from Y to Z. C Electrons move from X to W and from Z to Y. D Electrons move from X to W but not from Z to Y. Answer: C Electrons near W attracted to positive rod causing electrons to move towards W. X becomes net positive. X being positive attracts electrons to move towards Y, drawing electrons from the earth. 1.2 A 20 µF capacitor is charged by a constant current of 10 mA. If the capacitor is initially uncharged, how long does it take for the potential difference across the capacitor to reach 300 V? A 6.0 × 10–4 s B 0.60 s C 15 s D 6.0 × 105 s Answer: B Charge on capacitor when p.d. is 300 V, Q = CV = (20 × 10–6)(300) = 0.006 C Since current = charge / time, 10 × 10–3 = 0.006 / t t = 0.6 s

National Junior College Science Department | Physics 2 1.3 Four identical capacitors are connected as shown. Which of the following lists the arrangements in order of decreasing capacitance? A PQRS B PSRQ C QRSP D QSRP Answer: B Let C be the capacitance of each capacitor. Circuit P: Ctotal = C + C + C + C = 4C Circuit Q: Ctotal = [1/C + 1/C + 1/C + 1/C]–1 = C/4 Circuit R: Ctotal = [1/C + 1/(C + C) + 1/C]–1 = 2C/5 Circuit S: Ctotal = [1/C + 1/(C + C + C)]–1 = 3C/4 Decreasing capacitance is PSRQ. 1.4 Two capacitors are connected in series as shown. What is the charge carries by each of these capacitors? charge on C1 / μC charge on C2 / μC A 4.0 4.0 B 4.0 8.0 C 8.0 4.0 D 8.0 8.0 Answer: D Ctotal = (1/1.0 + 1/2.0)–1 = 2/3 μF p.d. across equivalent capacitor V = 12 – 0 = 12 V Charge on equivalent capacitor = Ctotal V = (2/3) (12) = 8 μC Capacitors in series have the same amount of charge.

National Junior College Science Department | Physics 3 1.5 The energy stored in a capacitor of capacitance C, carrying charge Q with potential difference V between its plates. may be obtained by calculating the area under an appropriate graph. Which graph shows the correct relationship between a pair of the quantities C, Q and V, and in addition shows a shaded area which corresponds to the energy stored in the capacitor? Answer: A C = Q / V and energy stored in capacitor =!"𝑄𝑉=!"𝐶𝑉"=!"#!$. So Option A: Graph of V against Q gives a straight line with gradient = 1/C. Area under graph = !"𝑄𝑉 = energy stored Option B: Graph of C against Q gives a straight line with gradient = 1/V. Area under graph = !"𝑄𝐶 ≠ energy stored Option C: Graph of C against 1/V gives a straight line with gradient = Q. Area under graph = !"𝐶'!%(=!"$% ≠ energy stored Option D: Graph of Q against V cannot be a horizontal line. 1.6 When the potential difference V between the plates of a capacitor is increased from V1 to V2, the charge Q on the plates increases from Q1 to Q2. On the graphs, which shaded area represents the increase in the energy stored in the capacitor? Answer: D Initial energy stored = ½ Q1V1 Final energy stored = ½ Q2V2 Increase in energy stored = ½ Q2V2 – ½ Q1V1

National Junior College Science Department | Physics 4 1.7 The diagrams show two ways of connecting two identical capacitors of capacitance C in circuits each with a supply of p.d. V. What is the value of total energy stored in the capacitors in circuit 1total energy stored in the capacitors in circuit 2 ? A 0.25 B 0.50 C 2.0 D 4.0 Answer: A Circuit 1: Equivalent capacitor Ctotal = (1/C + 1/C)–1 = C / 2 p.d. across equivalent capacitor = V Energy stored = ½ (C / 2) V2 = ¼ CV2 Circuit 2: Equivalent capacitor Ctotal = C + C = 2C p.d. across equivalent capacitor = V Energy stored = ½ (2C) V2 = CV2 Ratio = ¼ / 1 = 0.25 1.8 A capacitor is charged to a voltage V and then discharged through a small d.c. motor. As the capacitor discharges, the motor raises a mass through a height h. The experiment is repeated for several values of V. A constant fraction of the capacitor energy is converted to gain of gravitational potential energy. Which graph would be expected to give a straight line? A h against V 2 B h against V C h against √V D h against 1/√V Answer: A Energy stored in capacitor U = 12CV2 Energy transferred to gravitational potential store = k × U = mgh where k is a constant < 1. k×12CV2=mgh h=kC2mgV2 Since k, C, m and g are constant, a graph of h against V 2 will give a straight line.

National Junior College Science Department | Physics 5 1.9 In the circuit shown, a capacitor of capacitance 3 µF is charged from a battery of e.m.f. 6 V with the switch connected to terminal P. The switch is now connected to Q. This charges the 6 µF capacitor from the 3 µF one. What is the new potential difference across the combination? A 1 V B 2 V C 4 V D 6 V Answer: B Before connecting to Q, charge on 3 µF capacitor = (3)(6) = 18 µC After connecting to Q, equivalent capacitance of parallel connected capacitors = 3 + 6 = 9 µF charge on equivalent capacitor = 18 µC p.d. across equivalent capacitor V = Q / C = 18 / 9 = 2 V 1.10 A capacitor of capacitance C1 is charged to a potential difference of 100 V and then disconnected. When a second uncharged capacitor of capacitance C2 is connected in parallel across it, the new p.d. is 60 V. What is the value of the ratio C1C2? A 0.60 B 0.67 C 1.50 D 1.67 Answer: C Before connecting to C2 capacitor, charge on C1 capacitor = 100C1 After connecting to C2 capacitor, equivalent capacitance of parallel connected capacitors = C1 + C2 charge on equivalent capacitor = 100C1 p.d. across equivalent capacitor V = Q / C = 100C1 / (C1 + C2) = 60 V C1+C2100C1=160 1+C2C1=10060 C1C2=6040=1.50