CAPACITORS AND CAPACITANCE

UNIT 5 CAPACITORS AND CAPACITANCE 5.1 Capacitor and Capacitance 5.1.1 Associated Quantities - A capacitor is an electrical device that is used to store electrical energy. - The unit of capacitance is Farad. The symbol of capacitance is C. - Capacitance is defined to be the amount of charge Q stored in between the two plates for a potential difference or voltage V existing across the plates. In other words: - A capacitor has capacitance of one Farad when current charging of one Ampere flows in one second. This process causing a transferring of one volt in plates potential. - The Farad unit is too large for practical as charge ratio to its potential difference. uses. Thus microfarad (μf), nanofarad (nf) or Pico farad (pf) is used as a suitable unit for capacitor:- Microfarad (µf) 1µF = 1/1,000,000F = 10-6 F Nanofarad (nf) 1nF = 1/1,000,000,000 = 10-9 F Microfarad (pf) 1pF = 1/1,000,000,000,000 = 10-12 F 5.1.2 Types of capacitors Mica Unpolarised Ceramic Film Fixed Air-gap Paper Polarized Aluminum Tantalum Variable Trimmer MARLIANA/JKE/POLISAS/ET101-UNIT5 1

5.1.3 Capacitor Construction Figure 1: Capacitor construction In its most elementary state a capacitor consists of two metal plates separated by a certain distance d, in between the plates lies a dielectric material with dielectric constant =ε o ε, where ε o is the dielectric of air. The dielectric material allows for charge to accumulate between the capacitor plates. Air (actually vacuum) has the lowest dielectric value of ε o = 8.854 x 10-12 Farads/meter. All other materials have higher dielectric values, since they are higher in density and can therefore accumulate more charge. The Physical meaning of capacitance can be seen by relating it to the physical characteristics of the two plates, so that, the capacitance is related to the dielectric of the material in between the plates, the square area of a plate and the distance between the plates by the formula: Clearly, the larger the area of the plate the more charge can be accumulated and hence the larger the capacitance. Also, note that as the distance d increases the Capacitance decreases since the charge cannot be contained as 'densely' as before. By applying a voltage to a capacitor and measuring the charge on the plates, the ratio of the charge Q to the voltage V will give the capacitance value of the capacitor and is therefore given as: C = Q/V this equation can also be re-arranged to give the more familiar formula for the quantity of charge on the plates as: Q = C x V. MARLIANA/JKE/POLISAS/ET101-UNIT5 2

5.2 Capacitor equivalents circuit 5.2.1 Capacitor connected in series e1 e2 e3 Total voltage V T = e 1 + e 2 + e 3 Since then E C1 C2 C3 Figure 2: Capacitors in series Where C T is the total equivalent circuit capacitance It follows that for n series-connected capacitors: 5.2.3 Capacitor connected in parallel E C1 C2 C3 Figure 3: Capacitor in parallel Total charge, Q T = Q 1 + Q 2 + Q 3 C T E = C 1 V 1 + C 2 V 2 + C 3 V 3 Total voltage E T = e 1 = e 2 = e 3 Total equivalent circuit capacitance C T = C 1 + C 2 + C 3 It follows that for n parallel connected capacitors: C T = C 1 + C 2 + C 3 + + C n 5.2.3 Capacitor connected in series-parallel Total equivalent circuit capacitance E C1 C2 C3 Figure 4: Capacitor in series-parallel MARLIANA/JKE/POLISAS/ET101-UNIT5 3

Example 1 Calculate the equivalent capacitance of two capacitors of 3μF and 6μF connected: (a) in parallel (b) in series. Solution: (a) In parallel, equivalent capacitances: (b) In series, equivalent capacitance : Example 2 Find the capacitance to be connected in series with a 10μF capacitor for the equivalent capacitance to be 6μF. Solution:,, For two capacitance in series: Example 3 Find the total capacitance of the circuit: E C1 C2 2µF C3 C4 1µF 3µF 4µF Solution: MARLIANA/JKE/POLISAS/ET101-UNIT5 4

TUTORIAL 1 1. Capacitors of 15μF and 10μF are connected (a) in parallel and (b) in series. Determine the equivalent capacitance in each case. 2. Find the capacitance to be connected in series with a 25μF capacitor for the equivalent capacitance to be 10µF. 3. Find the capacitance to be connected in parallel with a 25μF capacitor for the equivalent capacitance to be 10µF. 4. Find the total capacitor the circuit. a) E C4 C2 C1 100µF 100µF 100µF C5 50µF C3 200µF [40µF] b) C1 C2 C3 C4 60µF 60µF C5 C6 20µF 40µF [11.14µF] c) C1 C2 C3 C4 C5 C6 [1.2µF] MARLIANA/JKE/POLISAS/ET101-UNIT5 5

5.3 Circuit with capacitive load Figure 5 Mathematically, the capacitance of the device relates the voltage difference between the plates and the charge accumulation associated with this voltage: q(t)=cv(t) equation (1) Capacitors which obey the relationship of equation (1) are linear capacitors, since the potential difference between the conductive surfaces is linearly related to the charge on the surfaces. Note that the charges on the right and left plate of the capacitor in Figure 5 are equal and opposite. Thus, if we increase the charge on one plate, the charge on the other plate must decrease by the same amount. This is consistent with our previous assumption electrical circuit elements cannot accumulate charge, and current entering one terminal of a capacitor must leave the other terminal of the capacitor. So, current is defined as the time rate of change of charge, 5.3.2 Elements related to capacitance a. Electric field: Area that surrounds the electric charge or charges system where the increasing and decreasing of electric force exists. b. Line of electric force: A line of electric force is known as line or curve that pointed out from positive charge (+) to negative charge (-) in a magnetic field. c. Electric flux: Known as amount of electric force line pointed out from positive charge (+) to negative charge (-) in a magnetic field. Flux symbol is Ψ(phi). MARLIANA/JKE/POLISAS/ET101-UNIT5 6

d. Electric flux density (D): Electric flux density is a measurement of electric flux that pass through a unit of plate s area with a coincide angle, that is an area of 1 meter². The ratio between the charge of the capacitor and capacitor plates. The symbol used is D. Based on Figure 1, if the area of capacitor is A, then the flux density is given as:- where Q= charge(coulomb), A = surface area of capacitor Figure6: Area and distance of capacitor plates e. Electric field strength: When two metal plates are charged and separated in a certain distance, a potential difference will exists between the plates. A force was also generated, known as electric force and the symbol is E. The magnetic strength depends on the potential difference and distance between plates. Where; V = potential difference d = thickness of dielectric f. Dielectric: Insulator that is used between the two plates of a capacitance is known as dielectric. Electric field exists in the dielectric and the flux density depends on the types of insulator used. g. Absolute permittivity (ε): Permittivity is a capacitance or ability to store energy of a capacitor. A force was also generated, known as electric force and the symbol. It depends on the dielectric substance, and the symbol is ε. MARLIANA/JKE/POLISAS/ET101-UNIT5 7

5.3.3 Factor that effecting capacitance Based on Figure 7, the factor that effecting capacitance is: Figure 7 a. Capacitance between two plates proportional to the surface area b. Capacitance between two plates inversely proportional to the thickness of dielectric c. Increasing the dielectric constant of the material between the plates 5.4 Process charging and discharging in a capacitor 5.4.1 Charging process in capacitor b i V R V C + - a S R C A + E - Figure 8: Capacitor circuit for charging process In initial state, a capacitor is uncharged (V c = 0V). When a capacitor start charged, maximum current will be flowing (i = I max ). The current would be decreased by exponent, while voltage will be rising by exponent also. This state will continue until full state (steady) achieved. In this full state, current had decreased to zero value, while voltage increased until maximum value. The capacitor is said in fully charge. MARLIANA/JKE/POLISAS/ET101-UNIT5 8

Figure 9 show both curves current and voltage for capacitor during charging process, in x-axis (t, time). The current and voltage curve may be represented by exponent equations respectively. v c, i I Max V Max A v c = 0.632 V Max Voltage through capacitor: v c = V max (1 e t/ ) i = 0.371 I Max Current flow: i = I max (e t/ ) v c = V max i = 0 Figure 9: Current and voltage curve in capacitor during charging process Time constant, = CR - The times taken for voltage achieve value of 0.632V max and current achieve value of 0.371I max Initial current, = CR 5.4.2 Discharging process in capacitor t b i V R V C + - a S R C + E - Figure 10: Capacitor circuit for discharging process A When capacitor fully charge and then switch being transformed to b, discharge process for capacitor will happen. The time taken to recharge and fully discharge is 5 =CxR. Figure 11 show the curve for discharging process in capacitor. MARLIANA/JKE/POLISAS/ET101-UNIT5 9

v, i V max Voltage through capacitor: v c = V max ( e t/ ) t, time Discharging current flow: I max i =-I max (e t/ ) Figure 11: Current and voltage curve in capacitor during discharging process 5.4.3 Energy stored in a capacitor During charging process through capacitor, it will get energy. Energy is kept in static form. The voltage in capacitor will increase from 0 volt to E volt. Example 1 One capacitor 0.326µF connected in series with 680kΩ and dc voltage 120V. Determine: Solution: i. ii. i. Time constant ii. Initial current charge iii. Current through capacitor, 100ms after charge to the source. iv. Energy stored in capacitor. iii. iv. MARLIANA/JKE/POLISAS/ET101-UNIT5 10

Example 2 S Solution: 100kΩ 150µF + - R C 220V When switch are connected, calculate: i. Time constant ii. Initial current charge iii. Time taken for voltage through capacitor increase to 160V. iv. Current and potential difference through capacitor, 4 second after charge to the source. v. Energy stored in capacitor. i. ii. iii. v c = 160V iv. 47s iv. Example 3 200kΩ + 20µF - R C S 150V When switch are connected, calculate: i. Initial current charge ii. Initial potential difference through capacitor. iii. Time constant iv. Time taken for capacitor fully charges. Solution: i. ii. iii. iv. MARLIANA/JKE/POLISAS/ET101-UNIT5 11

TUTORIAL 2 1. A 20µF capacitor is connected in series with a 50 kω resistor and the circuit is connected to a 20 V, d.c. supply. Determine: a) The initial value of the current flowing, b) The time constant of the circuit, c) The value of the current one second after connection, d) The value of the capacitor voltage two seconds after connection, e) The time after connection when the resistor voltage is 15 v [0.4mA,1s,0.147mA,17.3V,0.288s] 2. A circuit consists of a resistor connected in series with a 0.5µF capacitor and has a time constant of 12 ms. Determine: (a) The value of the resistor (b) The capacitor voltage 7 ms after connecting the circuit to a 10 V supply *24kΩ,4.42V] 3. An 12µF capacitor is connected in series to a 0.5MΩ resistor across the dc voltage supply of 240V. Determine: (a) Time constant (b) Initial charging current (c) Time for capacitor voltage increase to 150V (d) The current flowing through the capacitor after 4 seconds (e) Energy stored in the capacitor when it is fully charged (f) Sketch the current and voltage (IV) curve to show the process of charging the capacitor. [6s,0.48mA,5.88s,0.246mA,0.346J] 4. A capacitor is charged to 100 V and then discharged through a 50 kω resistor. If the time constant of the circuit is 0.8 s, determine: (a) The value of the capacitor, (b) The time for the capacitor voltage to fall to 20 v, (c) The current flowing when the capacitor has been discharging for 0.5 s (d) The voltage drop across the resistor when the capacitor has been discharging for one second. [16µF,1.29s,1.07mA,28.7V] 5. A 0.1µF capacitor is charged to 200 V before being connected across a 4 kω resistor. Determine: (a) The initial discharge current (b) The time constant of the circuit [50mA,0.4s] MARLIANA/JKE/POLISAS/ET101-UNIT5 12