Activity 3: Energy Stored in a Capacitor 1. A voltage of 12 V is placed on a capacitor with C = 85 pF. (a) What is the charge on the capacitor? (b) How much energy is stored in the capacitor? 2. In open-heart surgery, a much smaller amount of energy will defibrillatorhe heart. (a) What voltage is applied to the 7 F capacitor of a heart defibrillator that stores 60.0 J of energy? (b) Find the amount of stored charge. 3. Calculate the energy stored in a 5 uF capacitor charged to 30 V.Did you enjoy watching television during this time of pandemic? But you can enjoy it more if you can't hear any electrical noise from it. And that is what capacitors do in your television. Because capacitors hold electric charges. they act as dampers. slowing down the sudden movement of current. including noise. Without capacitors in the power supply. the television would have a noisy picture and a persistent low-pitched buzz in the speakers. Several capacitors can be connected to be used in a variety of applications. They can be arranged in two simple and common types of connections. knownas series and parallel. for which we can easily calculate the total capacitance. These two basic combinations. series. and parallel can also be used as part of more complex connections. The Series Combination of Capacitors As for any capacitor. the capacitance oi the combination is related to both charge and voltage: C=QIV When this series combination is connected to a battery with voltage V. each of the capacitors acquires an identical charge Q. To exp lain. rst note that the charge on the plate connected to the positive terminal of the battery is *0 and the charge on the plate connected to the negative terminal is 0. Charges are then induced onthe other plates so that the sum of the charges on all plates. and the sum of charges on any pair of capacitor plates. is zero. However. the potential drop in = :2- on one capacitor may be different from the potential drop Vznon another capacitor. because. generally. the capacitors may have different capacitances. The series combination of two or three capacitors resembles a single capacitor with a smaller capacitance. Generally. any number ofcapacitors connected in series is equivalent to one capacitor whose capacitance (calledthe equivalent capacitance] is smaller than the smallest of the capacitances in the series combination. Charge on this equivalent capacitor is the same as the charge on any capacitor in a series combination: That is. all capacitors ofo series combination have the some chums. 01=QI=02=QJ This occurs due to the conservation of charge in the circuit. When a charge Q in a series circuit is removed from a plate of the rst capacitor {which we denote as -Q].lt must be placed on a plate of the second capacitor [which we denote as +0]. and so on. Figure l: (a)ThmcapadmsarecmechedinsedesThemapimdeotdtechamemeadt place is Q. {B} The network. of capacitors in {a} is equivalent to one capacitor that has smaller capacitance than any oi'the individual capadlances In (a). and the charge on its places is Q. inmvmrmmWM-' WI. (Lamarnu https://phys.libretexts.org/@opt/deki/files/8157/CNX_UPhysics_25_02_Seriesjpg?revision=2 We can find an expression for the total (equivalent) capacitance by considering the voltages across the individual capacitors. The potentials across capacitors 1, 2, and 3 are, respectively, V1= "/CZ V2 = "/ Cz V3= "/ C3 9 These potentials must sum up to the voltage of the battery, giving the following potential balance: Vr = Vi+ V2+ V3+ ... Potential V is measured across an equivalent capacitor that holds charge Q and has an equivalent capacitance CS. Entering the expressions for V1, V2, and V3, we get Ci Cz C3 Canceling the charge Q, we obtain an expression containing the equivalent capacitance, CT, of three capacitors connected in series: CT Ci CZ C3 This expression can be generalized to any number of capacitors in a series network. Sample Problem Four capacitors with the capacitance of 4 uF, 3 UF, 6 UF, and 12 uF respectively are connected in series with a battery of 12.0 V. Determine the following: (a) Equivalent capacitance (b) Total charge (c) Individual voltage Solution: (a) Equivalent Capacitance (c) Individual Voltage CT 4 UF 3UF 6UF 12F (QT = Q1 = Q2 = Q3 = Q4) 1 3+4+2+1 Q1 14.4 uC CT 12 UF V1= = 3.6 V 10 4 UF 1 CT 12 ULF Qz 14.4 HC = 4.8V 12 UF V2 = Cz 3 ULF CT = 10 Q 14.4 HC CT = 1.2 HF V3= 3 = C3 6 F - = 2.4 V (b) Total charge 14.4 uC QT = CTVT V= = 12 1 = 1.2V QT = 1.2 x 10-6 F (12.0 V) Q7 = 1.44 x 10-5 C Checking QT = 14.4 AC VT = Vi+ V2+ V3+ V. 12 V = 3.6 V + 4.8V + 2.4V + 1.2V 10 12 V = 12 VThe Parallel Combination of Capacitors A parallel combination of three capacitors. with one plate of each capacitor connected to one side of the circuit and the other plate connected to the other side. Since the capacitors are connected in parallel, they all have the some voltage V across their plates. VTIVIIV3= Val-*- Figure 2: [a] Three capacitors are connected in parallel. Each capacitor is connected directlyto the battery. [in] The charge on the equivalent capacitor is the sum oithe charges on the individual ca pa citora. waxwoonuemmwruxovamammmu However, each capacitor in the parallel network may store a different charge. To nd the equivalent capacitance C1 of the parallel network. we note that the total charge Q stored by the network is the sum of all the individual charges: Qr= QI+QI+QI+--- {in the left-hand side of this equation. we use the relation 0 = L's-if. which holds for the entire network. CrV= C1V+ C2V+ Cop-Pm This equation. when simplied. is the expression for the equivalent capacitance of the parallel network of three capacitors: Cr= 51+ C1+ C3+m This expression is easily generalized to any number ofcapacitors connected in parallel in the network. Sample Problem Four capacitors with the capacitance of 4 uF. 3 all\". 6 pF. and 12 pi? respectively are connected in parallel with a battery of 12.0 '9'. Determine the following: {a} Equivalent capacitance {b} Total charge {:1 individual charges 12 (c) Individual charge Solution (VT = V1 = V2 = V3 = V4) (a) Total Capacitance CT = CI+ C2 + C3 + C+ Q1 = CIVI = (4 HF) (12V) = 48 HC CT = 4 HF + 3 HF + 6 HF + 12 /F Q2 = CzV2 = (3 HF) (12V) = 36 MC CT = 25 HF Q3 = C3V3 = (6 HF)(12V) = 72 AC Q4 = CAV4 = (12 MF) (12V) = 144 AC (b) Total Charge QT = CTVT Checking QT = 25 HF (12V) QT = Q1+ Q2 + Q3+ Q4 QT = 300 HC 300 HC = 48 HC + 36 HC + 72 HC + 144 AC 300 HC = 300 HC Energy stored in a simple capacitor PEcap = _ C(AV)? = The energy stored in a capacitor is potential energy. It can be extracted from the capacitor and transforms into other forms of energy or can be used to do mechanical work. Sample Problem A heart defibrillator delivers 500 J of energy by discharging a capacitor initially at 20,000 V. What is its capacitance? I. Given: III. Solution: PE = 500 J C= 2PE V = 20,000 V V 2 C =? 2 (5.0 x 103 J) C= (2.0 x 10* V)2 II. Formula: 1.0 x 103 / C = 4.0 x 10 VZ 2PE C= VZ C = 2.5 x 10-6 F C = 2.5 HF