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Transcript
Physics beyond 2000 Chapter 14 Capacitance Capacitor • A capacitor is a device which can store charges and thus electrical energy. • The most common capacitor consists of two conducting plates separated by a non-conducting material called dielectric (such as air, paper, plastic film etc.) • There are two terminals, one connected to high potential and the other to low potential. 1. conducting plates 2. dielectric - 3. terminals - Capacitor • Electrolytic capacitor: note the connection of the terminals. The - terminal must be at a higher potential than the - terminal. Other kinds of capacitors Charge storage of a simple capacitor • Charging a capacitor: store charges in a capacitor. battery + 2. Connect the + terminal of the capacitor to the + terminal of the power supply + 1. Prepare a d.c. power supply e.g. battery - - capacitor 3. Connect the terminal of the capacitor to the - terminal of the power supply Charge storage of a simple capacitor • Charging a capacitor: store charges in a capacitor. battery 1. Positive charges flow into the capacitor through its - terminal + + - 2. Negative charges flow into the capacitor through its - terminal - capacitor Charge storage of a simple capacitor • Charging a capacitor: store charges in a capacitor. battery + - I=0 +Q -Q + capacitor When the capacitor is fully charged, its potential difference is equal to that of the power supply. No more charges flow now. Charge storage of a simple capacitor • Charging a capacitor: store charges in a capacitor. 3. The charged capacitor can act like a d.c. battery. It can supply electric current. battery + - +Q -Q + capacitor 1. Disconnect the capacitor and the power supply. The capacitor now stores some amount of charges, positive on one plate and negative on the other. 2. Though there are equal amount of positive and negative charges in the capacitor, we still say that the amount of charge is Q. Charge storage of a simple capacitor • http://maxwell.ucdavis.edu/~electro/dc_circ uits2/chrg_capacitor.html Discharging a capacitor • Discharging a capacitor: let the charges in a capacitor to flow out through an external circuit and use up the electrical energy. 3. Connect the capacitor to the circuit +Q -Q + - 1. A charged capacitor 2. A circuit with a resistor R R Discharging a capacitor • Discharging a capacitor: let the charges in a capacitor flow out through an external circuit and use up the electrical energy. 1. Positive charges flow out of the capacitor from its + terminal. This is the direction of current I 4. Electrical energy is consumed in the resistor +Q -Q + - R 2. Negative charges flow out of the capacitor from its - terminal. 3. The amount of charges in the capacitor decreases. Discharging a capacitor • Discharging a capacitor: let the charges in a capacitor to flow out through an external circuit and use up the electrical energy. Q=0 1. The current stops when when there is not any charge in the capacitor. I=0 R Charging and discharging a capacitor • http://www.fed.cuhk.edu.hk/sci_lab/ntnujav a/rc/rc.html Charging and discharging a capacitor 1. The switch S can connect either X or Y. X Y 2. A center-zero milliammeter to indicate the direction of current S + - G C 4. Circuit for charging the capacitor when S is connected to X. R 3. Circuit for discharging the capacitor when S is connected to Y. Charging and discharging a capacitor X Y S _ - current I 0 G R C charging the capacitor time t discharging the capacitor Charging and discharging a capacitor current I time t 0 charging the capacitor +Q + -Q - • Charges are flowing into the capacitor. • The current decreases exponentially. • The current stops when the capacitor is fully charged. current I Charging and discharging a capacitor time t 0 +Q Discharging the capacitor + -Q - • Negative current indicates that charges are flowing out of the capacitor. • The current decreases exponentially. • The current stops when the capacitor is fully discharged (empty). Charge storage and potential difference • Charging a capacitor 1. Charges accumulate on the plates of the capacitor. 2. The potential difference (p.d.) V increases. 3. More electrical energy is stored in the capacitor. +Q -Q V Charge storage and potential difference +Q d -Q V • Consider a parallel-plate capacitor with area A and charge Q. • The p.d. across the plates is V. • The separation between the plates is d. Charge storage and potential difference +Q d -Q V • Find the electric field strength E between the plates. Q E A Charge storage and potential difference +Q d -Q V • The voltage between the plates depends on the separation d. V E.d Charge storage and potential difference +Q d -Q V More charges can be stored for a capacitor with large area, small separation, high permittivity and high potential difference. • Combine the above two equations Q V .d A A Q .V d Capacitance C • This is to describe how much charges a capacitor stores. • Definition: The charges stored per unit voltage applied to a capacitor. Q C V Unit: C V-1 or F, farad. 1 F = 1 10-6 F Example 1 • The capacitance of a capacitor is a constant. • The higher the voltage applied to the capacitor, the more charges the capacitor stores. • There is a limit of the voltage, break-down voltage. Capacitance of parallel-plate capacitor Q A A Q .V C V d d Capacitance of parallel-plate capacitor Smaller area A - C A - d • A must be the overlapping area. - http://micro.magnet.fsu.edu/electromag/java/capacitance/ Bigger area A - Example 2 The dielectric material is air. The permittivity is o = 8.9 10-12 C2 N-1 m-2 - - air Electrometer Electrometer • It is used to measure a potential difference Vin without drawing a large current from the external circuit. However it can only measure a voltage of 1V at most. • It can measure charge and small current. Electrometer • Measure voltage 1. Disconnect all the contact screws by turning them fully anti-clockwise. 2. Turn the short-read knob to the read position. 3. Note that the voltage should not be higher than 1V. Electrometer • Measure charge – Use the built-in capacitor C of 10-8 F. 1. Disconnect all the resistor contact screws by turning them fully anti-clockwise. 2. Connect the capacitor by turning the capacitor contact screw fully clockwise 3. Turn the short-read knob to the short position and then the read position. 4. Take the reading of input voltage Vin. 5. Calculate Q = C.Vin • Electrometer Measure current – Use the built-in resistor R of either 1010 or 1011 . 1. Disconnect the capacitor contact screw and one of the resistor contact screw by turning them fully anti-clockwise. 2. Connect one of the resistor by turning its contact screw fully clockwise 3. Turn the short-read knob to the short position and then the read position. 4. Take the reading of input voltage Vin. 5. Calculate Vin I R Spooning charge • Use an electrometer to measure the charges • Note that the spoon carries equal amount of charges each time. The charges stored in the capacitor increases linearly. The voltage thus also increases linearly. Example 3 • This provides a method to measure the permittivity of a dielectric. Investigate the capacitance of a parallel-plate capacitor • Use the electrometer to measure the charges stored in the capacitor. 1 • Verify that Q A and Q d Investigate the capacitance of a parallel-plate capacitor • Use the electrometer to measure the charges stored in the capacitor. • Insert a dielectric between the plates. Verify that more charges can be stored without change in p.d. Variable capacitor • The capacitance can be changed by changing the overlapping area between the plates. Capacitors with dielectrics • When we insert a dielectric between the plates, the capacitance C increases markedly. dielectric parallel-plate capacitor Capacitors with dielectrics Consider an isolated capacitor with charges Q unchanged. 1. The positive and negative charges of the molecules are displaced and form dipoles due to the electric field. 3. Only the charges on the surfaces of the dielectric have effect. + + + + + + + + + + + + +Q - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - -Q 2. The effect of charges inside the dielectric cancels. Capacitors with dielectrics Consider an isolated capacitor with charges Q unchanged. 1. Inside the dielectric, there is an electric field strength Eo due to the parallel plates + + + + + + + + + +Q ++ + - + - + - + - + - + - + - + - + - + - + - + - + 2. There is also an electric field strength E’ due to the dielectric dipoles. - + - + - + - + - -Q 3. The resultant electric field strength is Ed = Eo - Ed Capacitors with dielectrics Consider an isolated capacitor with charges Q unchanged. 1. The resultant electric field strength is Ed = Eo - Ed + + + + + + + + + + + + +Q - + - + E- d+ - + - E+ - + o - + - + - + - + - + - + - + - + - + - + - -Q 2. For a smaller electric field strength, V = E.d is reduced. 3. C = Q/V. So C increases when V is reduced. Use of dielectric • Advantages: 1. Keep the plates apart 2. Increase the capacitance 3. Reduce the p.d. and thus reduce the chance of electrical breakdown. Use of dielectric • Let Cd be the capacitance with dielectric of permittivity • Let Co be the capacitance with vacuum as the dielectric of permittivity o • The relative permittivity r = o Relative permittivity and breakdown field strength Material Relative permittivity εr Breakdown field strength E/Vm-1 Vacuum 1.0000 - Air 1.0006 106 Paper 2.7 107 Mica 5 2 × 108 Glass 7 3 × 108 Water 80 - Relative permittivity and breakdown field strength • What should be a good dielectric? 1. The relative permittivity is large. 2. The breakdown field strength is high. Electrolytic capacitor • This is a practical capacitor. • There may be leakage current. • Should not exceed its breakdown voltage. Or it will explode. Reed Switch 1. Two magnetic contacts M and N N M 3. An a.c. supply 2. One magnetic reed r r 5. A coil to magnetize the reed r. 4. A diode to allow current to flow in one direction only. Reed Switch N Currrent through the diode M T/2 current r time 1. Current (a.c.) flows through the coil and magnetize the reed r. The reed touches contact M. Reed Switch N Currrent through the diode M T current r time 1. Current is cut by the diode. The reed r touches the contact N. Reed Switch N N M r • The M current current r reed r contacts either N or M at a frequency f, equal to the frequency of the a.c. supply Measure C with a reed switch Measure C with a reed switch • Assume that the capacitor is fully charged and discharged in each cycle. The maximum charge stored in C is Q = C.V where V is the e.m.f of the battery Measure C with a reed switch • When the capacitor is discharged, the ammeter shows a current I, which is the average rate of flow of charge Q. I = Q.f where f is the frequency of the a.c. supply. Measure C with a reed switch • Q = C.V and • I = Q.f I • C= Vf In the experiment, read I from the milliammeter. V is the e.m.f. of the battery. f is the frequency of the a.c. power supply. Measure C with a reed switch • In the above calculation, the capacitor is assumed to be fully charged and discharged in each cycle. • In this case, the current I is independent on the resistor R in the circuit. • The resistor R is to protect the battery and the milliammeter. • In experiment, adjust R so that reducing R would not produce noticeable increase in current reading. Measure relative permittivity εr 1. Measure the capacitance Cd of the capacitor with the dielectric between the plates. Use reed switch for the measurement. 2. Measure the capacitance Co of the capacitor without the dielectrics. 3. Cd r Co Stray capacitance • The environment reduces the potential difference of a capacitor and increases the capacitance of a parallel-plate capacitor. • The measured capacitance C of the capacitor > expected capacitance Co. • The increase in capacitance is called the stray capacitance Cs. • C = Co + Cs Parallel-plate capacitor C C = Co + Cs Cs 0 1 d A where Co = d Parallel-plate capacitor C C = Co + Cs Cs 0 A where Co = A d Measure the voltage of a charged capacitor • What would happen if we use a moving-coil voltmeter to measure the voltage of a charged capacitor? 2. Connect a moving-coil voltmeter to the capacitor 1. A charged capacitor + + + + + + + - Measure the voltage of a charged capacitor • What would happen if we use a moving-coil voltmeter to measure the voltage of a charged capacitor? + + + + + + + - 3. The charges flow through the voltmeter and reduces the amount of charges in the capacitor Measure the voltage of a charged capacitor • What would happen if we use a moving-coil voltmeter to measure the voltage of a charged capacitor? + + + + + - 4. There is a leakage of charge Measure the voltage of a charged capacitor • The capacitor is discharged through the voltmeter. The reading on the voltmeter drops gradually. + + + + + - 4. There is a leakage of charge Measure the voltage of a charged capacitor • Use a voltage follower (will be studied later) to measure the voltage of the capacitor. • The voltage follower draws a very small amount of charges from the capacitor. Example 4 • Note that the charge Q on each plate is constant. • However the area A decreases when a plate is moved sideway. Prove that the p.d. between the plates decreases. Move sideway Capacitance of two concentric spheres 1. A small sphere with radius a and charge +Q at the middle 2. An earthed concentric shell with radius b. 3. What is the potential of the shell? V=0 Capacitance of two concentric spheres 1. By induction, there is charge on the inner surface of the shell. What is the charge? V=0 a +Q b -Q Capacitance of two concentric spheres E=? Electric field strength between the spheres V=0 V=? P.d. between the spheres a +Q b -Q C=? Capacitance of the concentric spheres Capacitance of two concentric spheres • Use Gauss’ law to find the electric field strength Er between the spheres. V=0 a +Q Er = ? b r -Q 1 Q Er . 2 4 r Capacitance of two concentric spheres • Find the p.d. V between the spheres. • Note that V = 0 at r = b. (Not ∞!) • Hint: V Er .dr V=0 a +Q V=? b -Q Q 1 1 V .( ) 4 b a Capacitance of two concentric spheres • Find the capacitance C. C=? V=0 a +Q b -Q ab C 4 ba Capacitance of an isolated sphere 1. A conducting sphere with radius a. 2. What is its capacitance C? 3. Put charge +Q on it V Q 4a C 4a a +Q 4. Imagine a shell with radius ∞ enclosing this sphere 5. Find the p.d. V and then C. Example 5 • Find the capacitance of earth. • Earth is an isolated sphere. Example 6 • If C = constant, ΔQ = C. ΔV. Combination of capacitors • In parallel : + - All positive terminals + are connected together. So are the negative terminals. + - + - • In series : The positive and negative terminals are in succession. + + - - + - + - - Capacitors in parallel C Ci Prove that C = C1 + C2 for two capacitors in parallel. C1 + Hint: Suppose that the p.d. across each capacitor is V. - + - + C2 Capacitors in series 1 1 C Ci Prove that parallel. 1 1 1 C C1 C2 for two capacitors in Hint: Suppose that the charge of each capacitor is Q. + +Q -Q +Q -Q - Example 7 • You may keep the unit μC in your calculation. Doubling voltage rating • If we have capacitors Co with breakdown voltage Vo , how to combine the capacitors so that they can stand 2.Vo? Co + - Use 4 capacitors. Co + - + Prove it. - + Co + Co - Doubling voltage rating • Note that the p.d. across each capacitor is only Vo. Vo Vo Co + - Use 4 capacitors. Co + - + Prove it. - + Co + Co - Combining capacitors 1. A capacitor C1 with charge Q. C1 +Q -Q C2 2. Two capacitors are combined in series. 3. Connect C1 to the combined capacitors. C3 4. Suppose that all capacitors have the same capacitance C. Combining capacitors • Charges in capacitor C1 will be shared with capacitor C2 and capacitor C3. C1 C2 +Q2 -Q2 +Q1-Q1 -Q3 C3 +Q3 Find the charges in each capacitor Combining capacitors • Compare the following two circuits. Find the relations among the charges. Note that charges are conserved! C1 +Q-Q C2 C1 C3 C2 +Q2 -Q2 +Q1-Q1 -Q3 C3 +Q3 Combining capacitors • Q1 + Q2 = Q Q2 = Q – Q1 • Q3 – Q2 = 0 Q3 = Q2 C1 +Q-Q C2 C1 C3 C2 +Q2 -Q2 +Q1-Q1 -Q3 C3 +Q3 Combining capacitors 1. 2. Q2 Q Q1 V2 C C V2 C2 Q1 V1 C V1 C1 +Q2 -Q2 +Q1-Q1 3. Q3 Q2 V3 V2 C C -Q3 +Q3 C3 V3 Combining capacitors The potential difference is zero round the circuit. V1 – V2 - V3 = 0 V1 = 2.V2 0V V1 C1 V2 C2 +Q2 -Q2 +Q1-Q1 -Q3 +Q3 C3 V3 Combining capacitors 1. 2(Q Q1 ) 2. C V1 for C2 V1/2 C2 Q1 C V1 for C1 V1 3. From 1 and 2, Q1 =2Q/3 and Q2 = Q3 = Q/3 C1 +Q2 -Q2 +Q1-Q1 -Q3 +Q3 C3 V1/2 Combining capacitors What is the capacitance of the combination? Hint: C2 and C3 are in series. C1 is in parallel with C2 and C3. V1 C1 V1/2 C2 Q 3 Q 3 2Q 2Q 3 3 Q 3 Q 3 C3 V1/2 Combining capacitors What is the capacitance of the combination? 3C C' 2 V1 C1 V1/2 C2 Q 3 Q 3 2Q 2Q 3 3 Q 3 Q 3 C3 V1/2 Energy stored in a capacitor Suppose that a capacitor of capacitance C stores charges Qo and the p.d. is Vo. V voltage V Vo Vo 0 1 .Q C 1 .Qo C Q charge stored Qo Energy stored in a capacitor If the charges has been increased by ΔQ, the work done is W V .Q Q .Q C The work done is represented by the area below the V-Q graph. V voltage V 1 .Q C Vo V 0 ΔQ Q charge stored Qo Energy stored in a capacitor The energy is 1 Qo2 1 1 U CVo2 QoVo 2 C 2 2 V voltage V 1 .Q C Vo 0 Q charge stored Qo Energy stored in a capacitor Another method is to find the sum of ΔW. Qo Q W W .Q Q 0 C Qo Q 1Q 0 C dQ 2 C V voltage 1 V .Q C Vo V 0 ΔQ Q charge stored Q 2 o Energy stored in an isolated conducting sphere QO a 1. Find the potential Vo. 2. Calculate the energy stored 1 from U = Vo Qo 2 3. Qo2 U . 8 a 1 Example 8 • Choose a suitable equation for the energy from The energy is 1 Qo2 1 1 2 U CVo QoVo 2 C 2 2 http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c1 Camera flash • C = 2000 μF and V = 50 V. • What is the energy stored in the capacitor? 2.5 J Camera flash • All the energy stored in the capacitor is given out in 1 ms. • What is the power of the flash? 2.5 kW Charging a capacitor with a cell S C emf Vo R empty capacitor Charging a capacitor with a cell S + I emf Vo R - C p.d.=V After closing the switch S, current I starts to flow in the circuit. Charges are stored in the capacitor. The current decreases with time when more charges are stored. The p.d. of the capacitor increases with time. Charging a capacitor with a cell S Qo ++++ I=0 emf Vo R C p.d. = Vo -Qo - - - - When the p.d. of the capacitor equals that of the cell, current stops. The capacitor is fully charged. Charging a capacitor with a cell S Qo ++++ I=0 C p.d. = Vo -Qo - - - emf Vo R 1. The electrical potential energy stored in C is 1 1 2 U C CVo QoVo 2 2 2. The electrical potential energy provided by the cell is Uo = QoVo = 2.UC Charging a capacitor with a cell S Qo ++++ I=0 C p.d. = Vo -Qo - - - emf Vo R Half of the energy is lost when current flows through the resistor R. Example 9 • The efficiency of the circuit is 50%. Typical examples • Energy stored in a capacitor is 2 1Q 1 1 2 U CV QV 2 C 2 2 V Q C V +Q -Q Joining 2 capacitors with same C 1. A charged capacitor with charge Q. Q + S C C - 3. Connect both capacitors by closing switch S. 2. An uncharged capacitor Q. S Joining 2 capacitors with same C Find the energy stored before the capacitors are connected. S 2 1Q Uo 2 C Q + C C - S Joining 2 capacitors with same C Find the energy stored after the capacitors are connected. S Q 2 + - C Q 2 S The charge in each capacitor is + C - Q 2 Joining 2 capacitors with same C Find the energy stored after the capacitors are connected. S Q 2 + - C Q 2 S The energy in each capacitor is + C - Q 2 ( ) 1 2 1 Uo 2 C 4 Joining 2 capacitors with same C Find the energy stored after the capacitors are connected. S Q 2 + - C Q 2 + C - S 1 The total energy is U o. Half of the energy is lost 2 in the wire. Inserting a dielectric with constant V • Fully charge a capacitor. The p.d. of the capacitor is equal to the emf of the cell, Vo dielectric C Vo Inserting a dielectric with constant V • Insert the dielectric between the plates. Note that the p.d. of the capacitor is still Vo. • The capacitance is C’ = εr.C dielectric C’ Vo Inserting a dielectric with constant V • As there is an increase in capacitance, the energy stored and charges stored also increase. 1 U ' C 'Vo2 Q ' C '. V o 2 dielectric Q’ C’ -Q’ Vo Inserting a dielectric with constant V • Current flows into the capacitor from the cell when the dielectric is inserted. • This increases both the energy and the charge in the capacitor. dielectric Q’ C’ -Q’ Vo Inserting a dielectric with constant Q • Charge the capacitor so that it stores charge Qo. dielectric Qo C -Qo Vo Inserting a dielectric with constant Q • Disconnect the capacitor from the cell by opening the switches. • Note that the charge is kept in the capacitor. dielectric Qo C -Qo Vo Inserting a dielectric with constant Q • Insert the dielectric between the plates. The capacitance C’ = εr.C. dielectric Qo -Qo C’ Vo Inserting a dielectric with constant Q • As there is an increase in capacitance, the energy stored and the p.d. decrease. 1 Qo2 U ' 2 C' dielectric Qo -Qo Qo V ' C' C’ Vo Inserting a dielectric with constant Q • There is not any current as Q is a constant. • Some energy of capacitor is lost to the dielectric. dielectric Qo -Qo C’ Vo Moving apart two charged capacitor with constant V Charge a parallel-plate capacitor. And then move the plates apart. Note that an external force is needed to do work. +Q C -Q Vo Moving apart two charged capacitor with constant V As the separation increases, the capacitance C’ is less. C’ Vo Moving apart two charged capacitor with constant V Both the energy and charge stored decrease. 1 U ' C 'Vo2 2 Q' C '.Vo +Q’ C’ -Q’ Vo Moving apart two charged capacitor with constant V There is current flowing from the capacitor to the cell when we pull the plates apart. This decreases the energy and the charge of the capacitor. +Q’ C’ -Q’ Vo Moving apart two charged capacitor with constant Q Charge a parallel-plate capacitor. The charge stored in the capacitor is Qo. +Qo -Qo C Vo Moving apart two charged capacitor with constant Q Off the switch to keep the charges in the capacitor. +Qo -Qo C Vo Moving apart two charged capacitor with constant Q Move the plates apart by an external force. The capacitance C’ is less. +Qo -Qo C’ Vo Moving apart two charged capacitor with constant Q The energy stored and the p.d. increase. 1 Qo2 U ' 2 C' Qo V ' C' Q’ C’ -Q’ Vo Moving apart two charged capacitor with constant Q The energy is increased due to the work done by external force to move the plates apart. Q’ C’ -Q’ Vo Charging a capacitor at constant rate • Keep the current constant Io while charging the capacitor. • Decrease R to keep the current constant. Read the current from the ammeter. Io 3. The current can maintain constant for a period of time S A 1. The p.d. of the capacitor increases and decreases the current. C Vo R 2. Decrease R from Ro to zero to increase the current. Charging a capacitor at constant rate • When the capacitor is full, the current stops suddenly. S A Qo C Vo -Qo R Charging a capacitor at constant rate • The charge in the capacitor is Qo and the p.d. is Vo. V Q I Vo Qo Io 0 to Vo Io Ro t 0 to Qo I o .to t 0 to Qo Vo C t Charging a capacitor at constant rate • The time required to fully charge a capacitor is to = C.Ro. It depends on the resistor and the capacitor. V Q I Vo Qo Io 0 to Vo Io Ro t 0 to Qo I o .to t 0 to Qo Vo C t Charging a capacitor at constant rate • In practice, there are other resistances. The current does not drop to zero suddenly. I Io 0 to t Charging a capacitor via a resistor • Close the switch to charge an empty capacitor. Vo • Initially (at t = 0), the current Io = . R Io S A C Vo R Charging a capacitor via a resistor • The current I decreases with time as the charge in the capacitor accumulates. I S A Q C Vo I R VR -Q Vc Charging a capacitor via a resistor • There are three p.d. (Vo, Vc and VR) Vo is constant I S A Q C Vo dQ VR IR R dt I R VR Vc -Q Q VC C Charging a capacitor via a resistor • Use Vo = VC + VR, to find the charge Q stored in the capacitor at time t. t Q Qo (1 e CR ) I S A Q C Vo I R VR -Q Vc Charging a capacitor via a resistor Q I Io Qo 0 t Q Qo (1 e t CR ) 0 t I I o .e t CR Charging a capacitor via a resistor What is the time t1/2 needed to fill the capacitor by half? Q 1 Hint: Q = Qo. 2 Qo 1 Qo 2 0 t1/2 = CR.ln2 = 0.693CR t t1/2 Q Qo (1 e t CR ) t1/2 is the half-time. The product CR is called the time constant. Charging a capacitor via a resistor • If there is not any resistor in the circuit, a large current pulse will flow through and the capacitor is fully charged almost immediately. I S A Q Vo C I -Q Vc Charging a capacitor via a resistor Vc VR Vo Vo 0 t V Vo (1 e t CR ) 0 t VR Vo .e t CR Examples 10 and 11 • Time constant = CR. Longer time is needed to charge the capacitor fully for a circuit with large CR. • Half-time t1/2 = CR.ln2. Discharging a capacitor via a resistor Suppose that the initial p.d. of the capacitor is Vo and the charge stored is Qo S A Q0 C Vc = Vo -Q0 R Qo = C.Vo Discharging a capacitor via a resistor The initial current Io = S A Vo Qo R CR Io Qo C Io R -Qo Vc = Vo Discharging a capacitor via a resistor The current I decreases with time. I I .e o S A I Q C I R VR -Q Vc t CR Discharging a capacitor via a resistor The charge Q decreases with time. Q Qo .e S A I Q C I R VR -Q Vc t CR Discharging a capacitor via a resistor Q I Qo Io 0 Q Qo .e t CR t 0 I I o .e t CR t Discharging a capacitor via a resistor Vc VR Vo Vo 0 VC Vo .e t CR t 0 VR Vo .e t CR t Discharging a capacitor via a resistor VC Vo .e t CR VR Vo .e t CR Discharging a capacitor via a resistor Find the time needed for the voltage to drop by half. Discharging a capacitor via a resistor • The time constant = CR. • The half-time t1/2 = CR.ln2 • Similar to that of charging a capacitor via a resistor. Energy lost by resistor during charging and discharging • UR = I2.R. t 2 Q UR = 2 o I . R . dt 0 2C • During charging, half the energy supplied by the battery is lost by the resistor. And half of the energy is stored in the capacitor. • During discharging, all the energy stored in the capacitor is lost. Example 12 • For a circuit with large CR, more time is needed to fully discharge the capacitor. Investigation with square waves 1. to CRO2 Use a signal VR generator to input a square wave (a.c.) R 2. Use CRO2 to view the p.d. VR with time It also shows the current I with time. High V I C Vc Low V 3. Use CRO1 to view the p.c. VC with time to CRO1 Investigation with square waves to CRO2 VR R High V I C Vc to CRO1 Low V Note that the two CRO cannot be connected at the same time in practice because of the ground lead. Investigation with square waves Input voltage 0 t 1. Charging time VC 0 t 2. Discharging time VR 0 t Investigation with square waves Input voltage 0 t VC 0 t VR 0 t If CR is too large, each period of input voltage is not enough for complete charging and discharging the capacitor. Investigation with square waves Input voltage 0 t VC 0 t VR 0 t If CR is too small, complete charging and discharging occur before the end of each cycle of the input voltage.