3/17/2009 PHYS202 SPRING Lecture notes Electric Circuits


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1 PHYS202 SPRING 2009 Lecture notes Electric Circuits 1
2 Batteries A battery is a device that provides a potential difference to two terminals. Different metals in an electrolyte will create a potential difference, and maintain the difference so long as the load on the battery is not to large. Chemical potential energy is converted to electric potential energy. The battery is said to provide Electromotive Force (E) 2
3 Ohm s Law Let us consider a circuit. Connections are shown as solid lines. Batteries are shown as two lines perpendicular to connections with one shorter than the other. The conventional current flows from the positive to negative terminals. Positive Negative I current R resistance 3
4 Ohm s Law Normally current is not labeled. The current flow is proportional to the voltage: Resistance is the constant of proportionality: Ohm s Law = I IR An element within an electric circuit with resistance is called a resistor. Ohm s Law is not a fundamental rule of nature. It isn t true in every case, everywhere but it is a very useful rule in practical situations. I current R resistance 4
5 Example 20.4 Suppose that the resistance between the walls of a biological cell is 5.0 x 10 9 Ω. (a) What is the current when the potential difference between the walls is 75 m? Use Ohm s law: = IR I = R I = = Ω A (b) If the current is composed of Na + ions (q = +e), how many such ions flow in 0.50 s? ΔQ I = Δ Q= IΔt Δt + 1Na IΔt N =Δ Q N = Na 1e e 11 ( )( ) A 0.50s 7 19 Na Na + N = = C + ions 5
6 Resistance Resistance is given in terms of an objects tendency to resist the flow of electrons ρ. Resistance increases with the length of the material and decreases with its cross sectional area. R = ρl A All materials have some resistivity, ρ,to differing degrees. 6
7 Resistance As expected, insulators have very high resistivity and conductors very low resistivity. Semi conductors fall in between and their resistivity depends heavily on the amount of impurities. 7
8 Resistance and Temperature As temperature changes, so does resistivity. In small temperature ranges, this is given by ( ) ρ = ρ 1+ α T T 0 0 Where the ρ 0 is the resistivity at temperature T 0, and α is the temperature coefficient of resistivity. 8
9 Example Problem A copper wire has a resistance of 38.0 Ω at 25 C and 43.7 Ω at 55 C. What is the temperature coefficient of resistivity? Solving ρ = ρ0 1+ α T T0 for α R Ω 1 R Ω α = = = C T T 55 C 25 C 0 ( ) ( ) 1 9
10 Electric Power Energy q q P = Δ Δ Δ P t = = Δ Δ t Δ t P= I The amount of power used in a portion of a circuit, is the product of the voltage drop and the current in that section of the circuit. Using Ohm s Law: = IR, P = I( IR) P = R = = R 2 P I R P 2 10
11 Example A piece of Nichrome wire has a radius of 6.5 x 10 4 m. It is used in a laboratory to make a heater that uses 4.00 x 10 2 W of power when connected to a voltage source of 120. Ignoring the effect of temperature on resistance, estimate the necessary length of wire. Looking up the resistance of NiChrome wire, we find ρ = 100 x 10 8 Ω m. We need to find the resistance of a wire with a particular radius: ρl ρl R = = 2 A π r We need the resistance needed to dissipate 400W by a voltage of 120: 2 P = R R = 2 P Thus, the length of the wire must be ρl π r = L = 2 π r P ρp ( 120) π ( m) L = = 48m 8 ( Ω m)( 400 W) 2 11
12 Example Tungsten has a temperature coefficient of resistivity of (C ) 1. A tungsten wire is connected to a source of constant voltage via a switch. At the instant the switch is closed, the temperature of the wire is 28 C, and the initial power delivered to the wire is P 0. At what wire temperature will the power that is delivered to the wire be decreased to ½ of P 0? If the source of the voltage is constant and temperature affects resistance, can R 2 2 compe the power we find P = and P =, thus what we are looking for is R the point where the resistance is twice the value at 28 C. Since the diameter and length of the wire are unchanged, we are simply looking for where the resistivity is doubled: ( T T ) ( T T ) 2ρ0 = ρ0 1+ α 0 α 0 = T = + T0 = + 28 C = 250 C 1 α C ( ) 12
13 Alternating Current Let us treat the voltage source as oscillating as a sine wave (see image to right) Normal AC voltage in the US has a frequency of 60 Hz (e.g. most of Europe is 50 Hz). f = 60 Hz So what is the current? Ohm s Law gives 0 I = = sin 2 R R I = I sin 2 ft 0 ( π ) ( π ft) The current oscillates in the same way as the voltage. ( π ) = sin 2 ft 0 13
14 AC Power Power is sin ( π ) P = I = I ft Average power is half of the peak power. Average Power Average Power is P I I = = 14
15 AC Power We define this level of voltage and current as the root mean square (rms) voltage and current. 0 I0 rms = Irms = 2 2 Since this is the average power delivered over one period, it equates to the power delivered in an AC circuit. We apply Ohm s law = IR P = I In the US, we have rms = 120 (240 for heavy duty), which equates to a maximum voltage of 170 (340) rms rms 2 rms 2 P = P = IrmsR R 15
16 Example A light bulb is connected to a wall socket. The current in the bulb depends on the time t according to the relation I = (0.707 A) sin [(314 Hz)t]. (a) What is the frequency of the alternating current? We know that I = I 0 sin(2πft), so, f = 314 Hz 50 Hz 2π = (b) Determine the resistance of the bulb s filament. Using Ohm s law, = I R rms rm s R = = I / 2 rms I0 ( ) R = = 240Ω A (c) What is the average power delivered to the light bulb? rms ( ) rms P= rmsirm s = = = 60 W R 240Ω 16
17 AC Power Transmission Note that power transmitted through a power line is Let us say that 10 km of wire has a resistance of 1.7 Ω. If a power line is to transmit 10kW, is it better to have high voltage or high current? When dealing with power loss through a wire, use P = I 2 R. P = I rms rms 2 rms 2 P = P = IrmsR R oltage Current Power loss 1 k 10A 170W (1.7%) 10k 1A 1.7W (0.017%) A 1.7kW (17%) 17
18 Power loss in a wire Current must stay constant through a wire, or charge would start to build up. The oltage drop is continuous, so we would have to calculate how much power is lost at each point in the wire due to the constant drop in voltage. 18
19 Wiring Resistors in Series Take two resistors (R 1 and R 2 ) and place them in series. 1 2 R 1 R 2 Since charge does not build up, the current through both must be the same, so each must experience a voltage drop: Series ( ) = + = IR + IR = I R + R ( ) ( ) IR = I R + R R = R + R 1 2 Series 1 2 Resistors in Series add their resistances together to get the equivalent resistance. R = R + R + Series
20 Wiring Resistors in Series Power for resistors in series Find the current in the circuit: 1 2 The corresponding voltage drops across each resistor are: 1 = IR1 and 2 = IR2 2 2 The power dissipated is P = I R and P = I R I = R + R P = R and P = R ( R + R ) ( R + R ) R 1 R 2 The total power dissipated is as expected P = P + 1 P = 2 R R 2 2 P = I R + R = ( R + R ) ( R + R )
21 Examples The current in a 47Ω resistor is 0.12 A. This resistor is in series with a 28Ω resistor, and the series combination is connected across a battery. What is the battery voltage? For resistors in series, the current must be the same, so, ( ) ( )( ) = I R1+ R2 = 0.12A 47Ω+ 28Ω = 9 21
22 Examples Two resistances, R 1 and R 2, are connected in series across a 12 battery. The current increases by 0.20 A when R 2 is removed, leaving R 1 connected across the battery. However, the current increases by just 0.10 A when R 1 is removed, leaving R 2 connected across the battery. Find (a) R 1 and (b) R 2. Examining the series situation, we find that I 1 = I s +0.20A and I 2 = I s +0.10A: = Is ( R1+ R2) = I R = I A R = I R = I A R Thus, ( ) ( ) 1 1 s s 2 R = R = ( I A) ( I A) 1 2 s = Is + Is + Is + 2 ( 0.10 A)( 0.20 A) = 2Is + ( 0.30A) ( Is A) ( Is A) 2 2 I + ( 0.30A) I A = 2I + ( 0.30 A) I s s s s 2 Is = A s I s 22
23 Examples (cont.) Thus, I = A, so substituting back, R s ( 0.20 A) ( 0.10A) = I + R = I + R 1 2 Is s 1 s = = R = = A A A I A A A R = 35Ω R = 50Ω 1 2 s Note that this is not the only method to get to the solutions. 23
24 Wiring Resistors in Parallel Take two resistors (R 1 and R 2 ) and place them in parallel. R 1 R 2 The battery provides the same voltage to each resistor: R 1 R 2 24
25 Wiring Resistors in Parallel Take two resistors (R 1 and R 2 ) and place them in parallel. The battery provides the same voltage to each resistor. I 1 R 1 I 2 R 2 The currents must add up to the total current through the battery, 1 1 I = I1+ I2 = + = + R1 R2 R1 R = + = + Rparallel R R R R R 1 2 parallel 1 2 For resistors in parallel, the net resistance is the inverse of the sum of the inverse of the resistances = + + R R R parallel
26 Wiring Resistors in Parallel Power for resistors in parallel: Find the current through the resistors: I1 = I2 = R R 1 2 I 1 R 1 I 2 R 2 Power is P = I P1 = I1 P2 = I2 The total power given from the battery is P = P = R1 R2 1 1 P= P1+ P2 = + = + = R1 R2 R1 R2 R parallel 26
27 Examples The drawing shows a circuit that contains a battery, two resistors, and a switch. What is the equivalent resistance of the circuit when the switch is (a) open and (b) closed? When the switch is open, no current flows through R2, thus the equivalent resistance is 65.0Ω. When the switch is closed the resistors are in parallel, so the equivalent resistance is, RR 1 2 = + Rparallel = Rparallel R1 R2 R1+ R2 ( 65.0Ω)( 96.0Ω) Rparalle l = = 38.8Ω 65.0Ω Ω What is the total power delivered to the resistors when the switch is (c) open and (d) closed? 2 2 ( 9.00 ) Pop = = = 1.25 W R 65.0Ω P cl ( 9.00 ) ( 9.00) = + = + = 2.09 W R R 65.0Ω 96.0Ω 27
28 Examples A cylindrical aluminum pipe of length 1.50 m has an inner radius of 2.00 x 10 3 m and an outer radius of 3.00 x 10 3 m. The interior of the pipe is completely filled with copper. What is the resistance of this unit? As the hint in the book suggests, we can treat this as two parallel resistors. We need to find the resistance of each. The cross sectional areas are ACu = πrin AAl = π( Rout Rin ) Thus the resistances, CuL CuL AlL AlL RCu = ρ = ρ R 2 Al = ρ = ρ 2 2 ACu π Rin AAl π ( Rout Rin ) Placing the resistances in parallel, ( AlRin + ρcu ( Rout Rin )) 2 2 ( Rout Rin ) π R π in = + = + Rnet RCu RAl ρcul ρall πρ R + πρ R R = + = Rnet RCu RAl ρcuρall ρ ρ L Rnet π ρ ( ) Al in Cu out in Cu Al 3 = = Ω 28
29 Resistors in Series and Parallel Circuits can be wired with resistors in series and parallel. To find the current and power delivered by a battery, create equivalent circuits by combining resistors: Each circuit is equivalent from the point of view of the battery. Deal with each part of the circuit where you are combining two resistors, then move to the next. R 2 R 1 R 3 R 1 R 3 R p R = R + R S = + R R R p 1 S 29
30 Examples Find the equivalent resistance between points A and B in the drawing. We must combine the resistors until only one remains. Starting with the three resistors in series on the right, we find three resistors in series that can be combined: 2.00 Ω 6.00 Ω 4.00 Ω 3.00 Ω 6.00 Ω R = 1.00Ω+ 2.00Ω+ 3.00Ω= 6.00Ω S Now, there are two resistors in parallel on the right, combining those we get, 2.00 Ω 6.00 Ω 4.00 Ω 2.00 Ω = Ω 6.00Ω R p R = 2.00Ω p 30
31 Examples (cont.) Now, we have a pair of resistors in series, combining, 2.00 Ω 4.00 Ω R Ω S = Ω+ Ω= Ω We now combine the two resistors in parallel, 2.00 Ω 2.67 Ω = + R 4.00Ω 8. 00Ω p = 2.667Ω R p We can now combine the final resistors which are in series, R = 2.00Ω Ω= 4.67Ω S 31
32 Examples Three identical resistors are connected in parallel. The equivalent resistance increases by 700Ω when one resistor is removed and connected in series with the remaining two, which are still in parallel. Find the resistance of each resistor. Let us define the resistance in parallel = + + = R1 = R R R R R 1 1 R 3 Removing one resistor and placing it in series with the other two: = + = Rp R R R R 1 3 R2 = R+ R= R 2 2 = R + 700Ω 2 1 Substituting our equations for R 1 and R 2 into the above, R= R+ 700Ω R= 700Ω R = 600Ω 32
33 Internal Resistance of a Battery One way a battery is rated is by its voltage between its terminals. However, when a battery is connected to a load and there is current flowing, there is some internal resistance to a battery. The net result is to reduce the voltage between the terminals. The resistance inside the battery is constant, but the voltage drop of the terminals depends on the load: = I( R+ R int ) int The voltage across the terminals is equivalent to the voltage over resistor R. = R term R R ( + ) + R 33
34 Terminal oltage and Internal resistance For internal resistance of 0.01Ω 34
35 Terminal oltage and Current Through the Battery The more current through the battery, the lower the terminal voltage. If a 12 battery has an internal resistance of 0.010Ω and a current of 10A flowing through the battery. The terminal drop is If the current is 100A, then ( ) term = IRint = 12 10A 0.01Ω = 11.9 term ( ) term = IRint = A 0.01Ω = 11.0 term 35
36 Kirchhoff s Rules Sometimes circuits are too complex to analyze by reducing parallel and series resistors. For this we use two concepts: 1. Conservation of charge implies conservation of current. Let us assume that the current flows in the circuit as drawn. For conservation of current, the current flowing into a junction must equal the current flowing out. 36
37 Kirchhoff s Rules Take junction A. The current in must equal the current out. Label each branch of current. I left I middle A B I right Find the currents in an out of each point: Point A: Iright = Imiddle + Ileft Point B: I + I = I middle left right Note that both equations in this case are equivalent. For conservation of current, the current flowing into a junction must equal the current flowing out. 37
38 Kirchhoff s Rules Iright = Imiddle + Ileft A 2. Next we consider loops. I left I middle I right Here there are two loops. B Just like walking up and down hills, if you go in a loop the place you start/end must be at the same elevation. So, if you make a loop in a circuit, you come back to the same voltage (potential). This is Kirchhoff s Loop rule. The sum of potential rises and drops around any loop must be zero. 38
39 Kirchhoff s Rules Iright = Imiddle + Ileft Go around each loop noting the potential rises and falls. I left I middle A I right Left loop: Let s start at point A and go counter clockwise: 5 I 2Ω I 3Ω = 0 left ( ) ( ) Right loop, again starting at A, but let s go clockwise: 4Ω + 12 = 0 left ( I right )( ) ( ) If we went counter clockwise, we would get the same answer! 12 4Ω = 0 I right ( ) This is Kirchhoff s Loop rule. The sum of potential rises and drops around any loop must be zero. B 39
40 Kirchhoff s Rules Iright = Imiddle + Ileft ( ) Ileft ( ) ( ) 5 I 3Ω 2Ω = 0 left 12 4Ω = 0 I right We can now find all the currents as we have three equations and three unknowns. This is an especially easy case as there is only one unknown in each loop equation. I left I right = 3.0A I left = 1.0A I mid = 2.0A I middle Because the current is positive, it moves in the direction assumed. A B I right 40
41 Examples Two batteries, each with an internal resistance of 0.015Ω, are connected as in the drawing. In effect, the 9.0 battery is being used to charge the 8.0 battery. What is the current in the circuit? For this we only need to apply Kirchhoff s Loop Rule. Taking the current as flowing in the counter clockwise direction and starting with the upper right corner: ( ) I( ) 8.0 I 0.015Ω 0.015Ω 9.00 = I = 0 I = 33A The negative sign means that our original assumption about the current direction is wrong. So we have 33A going clockwise. 41
42 Examples Find the current in the 4.00Ω resistor in the drawing. Specify the direction of the current. We have three loops (one of which is the sum of the other two), and three branches with two junctions (one of which is redundent). Let us label the currents as in the lower drawing. Thus the junction rule at B gives, I1+ I3 = I2 The loop rules for the left loop starting a D and going clockwise gives, ( ) ( I )( ) 3 I 2.00Ω Ω = The loop rules for the left loop starting a E and going clockwise gives, ( ) I ( ) 9 I 4.00Ω Ω = A D Ω B + I 1 I I 3 C 4.00 Ω Ω E F
43 Examples A Ω B Ω E (cont.) I + I = I (1) I (2) I3 = 0 9 4I (3) 3 6 8I2 = 0 We have three equations and three unknowns. D We are only required to solve for I 3, so, substituting (1) into (3), ( ) 9 4I 6 8 I + I = I + 8I = I 1 I Ω + C Multiply equation (2) by four, 36 8I + 16I = 0, then adding to the above eqn., 1 3 I 3 F I = 0 I 3 3 = 1.82A The negative sign indicates that the current is in the opposite direction from our initial assumption. 43
44 Examples For the circuit shown in the drawing, find the current I through the 2.00Ω resistor and the voltage of the battery to the left of this resistor. Let us define a current I 1 flowing through the middle wire as shown. The junction rule gives, I + I 1 = 3.00 A The top loop gives (going counter clockwise), 24 I1 ( 6.00Ω) ( 3.00 A)( 4.00Ω) ( 3.00 A)( 8.00Ω ) = I1 36 = 0 I = 2.00A Substituting into the loop rule equation we get 1 I 1 I = 5.00A 44
45 Wheatstone Bridge Used to find resistances precisely. B R 3 R x R x is unknown but R 1, R 2 and R 3 are accurately known. A G C R 3 is a variable resistor. R 1 D R 2 Method is to vary R 3 until the galvanometer does not deflect (i.e. points B and D are at the same voltage. At that time, the voltage drops across R 3 and R 1 are equal: I 1 R 1 = I 3 R 3 45
46 Wheatstone Bridge I 1 R 1 = I 3 R 3 B R 3 R x So, it must be true that I 2 R 2 = I x R x A G C Since there is no voltage between B and D, I 1 = I 2 and I 3 = I x. R 1 D R 2 I 1 R 2 = I 3 R x Let s divide the two equations. Or, IR I R R R = = I R I R R R x 2 R x R2 = R 1 R 3 x R 3 is very sensitive, so the galvanometer switch is closed only briefly to check if the current is zero. 46
47 How does the Galvanometer Work? A DC Galvanometer measures current by using a coil spring, magnets and a loop of wire (more when we reach chapter 21). The rotation of the loop of wire gives the current passing through the system on some calibrated scale. The full scale current sensitivity of a galvanometer is the current required to move the needle (i.e. coil) deflect full scale. Depending on how the galvanometer is configured with resistors determines whether it reads current or voltage. 47
48 Measuring Current Ammeter Galvanometers can measure small currents. If the sensitivity is I m, it can read currents from about 0.02 I m up to I m. (e.g. if I m = 50 μa then the scale is 1 μa to 50 μa). Galvanometers (like batteries) have small internal resistances. If we want to measure current on a larger scale, a galvanometer is placed is parallel with a shunt resistor. Since the voltage drop across the galvanometer and resistor are the same: I R= I R R G G R G 48
49 Measuring Current Ammeter R = I R I R = I R R G G So, G G I R R If the resistance of the galvanometer is known, then we can set R to be whatever resistance needed to make I R the scale of the ammeter. G For example, if we want to measure up to 1 A, but the galvanometer can only take 50 μa and has an internal resistance of 30Ω, then we need I R to be A. R ( A)( 30Ω) 3 = = Ω A 49
50 Measuring oltage oltmeter A voltmeter also consists of a galvanometer and a resistor (usually large), but in series. G R Imagine we want to measure a full scale of, then we would have m ( ) = I R+ R R = R I For example, if we want the full scale to be 15 (again let s take I m = 50 μa and R G = 30Ω A m 15 R = 30Ω= 300 k Ω G G The internal resistance of the galvanometer is negligible. 50
51 Examples Two scales on a voltmeter measure voltages up to 20.0 and 30.0, respectively. The resistance connected in series with the galvanometer is 1680Ω for the 20.0 scale and 2930Ω for the 30.0 scale. Determine the coil resistance and the full scale current of the galvanometer that is used in the voltmeter. Using Ohm s Law, Subtracting one equation from the other, Solving for I m, ( ) ( ) = I R + R = I R + R 1 m 1 G 2 m 2 G I m = I R I R 2 1 m 1 m Ω 1680Ω I = 0.008A 2 1 = = R1 R2 m Substituting back into either equation yields R G = 820Ω 51
52 Examples A galvanometer with a coil resistance of 12.0Ω and a full scale current of ma is used with a shunt resistor to make an ammeter. The ammeter registers a maximum current of 4.00 ma. Find the equivalent resistance of the ammeter. Remember that the net current, is the current through the ammeter and the galvanometer: I = I + I I = I I = 3.85 ma The voltage across the galvanometer and shunt resistor is the same, R s s G IR s s G G 3 IG A = RG = 3 12 Ω Is A R = 0.468Ω s = I R ( ) The equivalent resistance is that of the shunt resistor and galvanometer coil in parallel: = + Req = 0.450Ω R R R eq s G s G 52
53 Capacitors as elements in a Circuit A Capacitor can be an element in a circuit, like a battery or resistor. It is denoted as or Since, q = C, the voltage of a capacitor in a circuit depends on the capacitance and the charge: q = C So, how does it work in a circuit? 53
54 Capacitors in Parallel and in Series Just as for resistors, when capacitors are in parallel, they see the same voltage, so the capacitors will charge: So, we get, q1+ q2 + q3 = C 1 + C2 + C 3 ( ) q = C = C + C + C tot P C 1 C 2 C 3 C p For capacitors in parallel: P ( ) C = C + C + C
55 Capacitors in Parallel and in Series Just as for resistors, when capacitors are in series, they sum of the voltage drops must equal the net voltage of the battery. So the C 1 C 2 C 3 capacitors will charge: = Since, the wire in between the capacitors begins as neutral, the charge must split evenly, so all capacitors are charged to the same level. q q q q = + + C C C C S = + + C C C C tot C S 55
56 Capacitors in Parallel and in Series The rules for adding capacitors and for adding resistors are reversed: Series Parallel Resistors R = R + R + R + S = R R R R P Capacitors = C C C C S C = C + C + C + S
57 Charging and Discharging Capacitors in a Circuit If a Direct Current circuit contains a battery, resistor, an initially uncharged capacitor and a switch, then the capacitor will take time to charge: / ( ) q = q e cap 0 1 t RC Where e is Euler s constant (e= ) q 0 =C 0, where 0 is the voltage of the battery. The capacitor will reach full charge in infinite time (although in reality we can set a percentage of full charge to get an acceptable finite time). 57
58 Charging and Discharging Capacitors in a Circuit If a Direct Current circuit contains a resistor, an initially charged capacitor and a switch, then the capacitor will take time to discharge: qcap = q e t/ RC Where e is Euler s constant (e= ) 0 q 0 is the initial charge on the battery. The capacitor will discharge in infinite time (although in reality we can set a low percentage of full charge to get an acceptable finite time). 58
59 Charging Capacitor Discharging Capacitor Charging and Discharging Capacitors in a Circuit q cap / ( ) q = q e cap 0 1 t RC q e t/ RC τ =RC, is the time constant of the circuit(s). When RC is small, the capacitor charges/discharges quickly When RC is large, the capacitor will charge/discharge slowly. = 0 59
60 Examples The circuit in the drawing contains two resistors and two capacitors that are connected to a battery via a switch. When the switch is closed, the capacitors begin to charge up. What is the time constant for the charging process? Find the equivalent resistance and capacitance: CS = C1+ C2 = + RP R1 R CS = 3.0μF+ 6.0μF = + RP 4.0kΩ 2.0kΩ C = 9.0μ F R = 1.333kΩ S p The time constant is the product of equivalent resistance and equivalent capacitance 3 τ = RC = Ω F p S ( )( ) τ = s 60
61 Examples How many time constants must elapse before a capacitor in a series RC circuit is charged to 80.0% of its equilibrium charge? Use the Charging equation: We want q cap to be 0.8q 0, 0 0 ( ) / ( ) q = q e cap 0 1 t RC t / τ ( ) 0.8q = q 1 e t = = τ t = ln 0.2 τ = 1.61τ ( ) t / τ 0.2 e ln
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