LCTRC CURRNT CHAPTR 25 LCTRC CURRNT and DRCTCURRNT CRCUTS Current as the motion of charges The Ampère Resistance and Ohm s Law Ohmic and nonohmic materials lectrical energy and power ons lectrons nside a conductor the charges are in constant, random motion. But generally, in metals, only the electrons are free to move, at speeds typically ~ 10 6 m/s. However, with no applied electric field the electrons move randomly: P 1 Simple circuits Combinations of resistors Series and parallel Kirchoff s Rules xample of an RC circuit P 2 Since there are many free electrons and their motions are uncorrelated, there is NO NT FLOW N ANY DRCTON.
V ab f a potential difference is applied, e.g., from a V b V a battery, a net flow of charge results... WHY? Because an! V electric field is setup a > V b and each electron experiences a force: F! = q! towards the right... P 2 P 1! F = e! As a result, an electron that would have travelled from P 1 P 2, say, now travels from P 1 P 3. ndeed, all the electrons experience a force to the right and so there is a net flow (or drift ) of ve charge from left to right. The drift velocity of electrons is typically ~ 10 4 m/s. P 3 V b The current is defined as the rate of flow of charge, i.e., the amount of charge passing through a given area! each second, i.e., = dq dt The direction of the current is defined as the direction ve charges would flow, i.e., in the same direction as! (that is from higher potential to lower potential). n our example, even though electrons flow from left right, the current direction is from right left. Note: current is a BULK effect (i.e., not restricted to the surface). UNTS: V a > V b V a dq Coulombs (C) dt seconds (s) C s Ampères (A)
f the charges drift an average distance Δx in a time Δt, the drift velocity v d is: v d = Δx Δt, i.e., Δx = v dδt. The number of charges in the shaded volume is: n(a.δx) where n is the number of charges per unit volume. n time Δt the amount of charge, ΔQ, that flows out of (and into) the volume is: so the current is: v d Δx ΔQ = n(a.δx)q = nqav d Δt, = ΔQ Δt = nqav d Δt = nqav d. Δt f the current is due only to electrons (normal in metals): = n e A v d. Area = A When there are two types of charges, e.g., in a fluorescent tube, the total current is: = A i n i q i v di = n 1 q 1 Av d1 n 2 q 2 Av d2, i.e., it is the sum of the currents due to each type of charge. NOT: v d 1 v d2 if q 1 is positive then v d1 > 0, i.e., in the same direction as the current, so q 1 v d1 > 0. if q 2 is negative then v d2 < 0, i.e., in the opposite direction to the current, so q 2 v d2 > 0. Therefore, the individual currents still add.!
Question 25.1: n a certain electrical device, 1C of charge flows from left to right and 5C of charge flows from right to left each second. 1C 5C 1C 5C A current of 5 C/s to the left is equivalent to a current of 5 C/s to the right, so the net current is What is the net current,? 1 C / s 5 C/s = 6 C/s = 6A.
OHM s LAW (1827) Specific resistance (also known as the resistivity): l " l Area A V V V b a > V b a The potential difference between sections a and b is: V ab = V a V b = l (V a > V b ) Georg Simon Ohm found that the ratio V ab was constant. The constant is called the resistance of that segment of wire, i.e., V ab = R. quivalent expressions for Ohm s Law are: UNTS: V ab = R and = V ab R. V Volts (V): Ampères (A) then R Ohms (Ω) The actual resistance of the section depends on the length l, the cross sectional area A and the type of material: R = ρ l A, where ρ is called the specific resistance or resistivity of the material. Typical values of ρ (in Ω.m) Silver 1.6 10 8 Copper 1.7 10 8 Tungsten 5.5 10 8 Nichrome 100 10 8 Carbon 3500 10 8 Glass 10 10 10 14 Rubber 10 13 10 16
ρ A ρ B ΔV Δ Ohmic (most metals) slope = Δ ΔV = 1 R constant ρ T T V Resistor T c C T The resistivity ρ of a material generally varies with temperature: A: typical metal B: typical semiconductor C: typical superconductor V However, not all materials and/or devices are Ohmic, i.e., obey Ohm s Law; look at the semiconductor diode below. Diode V Non Ohmic Consequently, Ohm s Law is only valid provided the temperature of the material is constant. V
Work done and power: lectrical power and energy: b l a UNTS: P Joules s watts (W). " δq 1kW = 1000W R V b V a V b V a Consider a charge δq moving under the influence of the electric field from a b in a time δt. The work done by the electric field is: δw = δq(v a V b ) = δq.v, where V is the potential difference between a and b, ( V a > V b ). So, the rate the work is done (i.e., the rate at which the system loses energy), i.e., the power, is: P = δw = V δq δt δt = V (Watts.) Also, using Ohm s Law: P = 2 R = V2 R. P = V = 2 R = V2 R V = (V a V b ) For charges to flow for t seconds, the energy required is: U = Vt = 2 Rt = V2 R t (Joules), which is often dissipated in the form of heat or light, e.g., Hairdryer: 1500W Light bulbs: 25W, 60W, 100W, etc. Note: You pay FPL for the work done in moving charges around circuits, which appears as the energy produced ( U = P t).
xamples of simple electrical circuits... Reflector Bulb Switch contact Switch Magnet Batteries Spring contact Rating: 60W at 120V Since P = V2, the resistance of the filament in the bulb R is R = V2 P = 1202 60 = 240Ω. Note: a lemon with copper and zinc nails acts as a battery! ach circuit has a battery and a load (bulb or motor). The batteries provide the potential difference to move charges through the load. Note: if a potential difference of only 100V is used, the resistance of the filament, R, doesn t change, but the power does: P = V2 R = 1002 240 = 41.7W, i.e., the power dissipated is less so bulb is not as bright. Copper nail Lemon Zinc nail
More on simple circuits... Flashlight with two batteries Switch open LOAD Switch closed Another example of a simple circuit; a 0.47Ω resistor is connected across the terminals of a 9V battery. The battery provides the potential difference to move ve charge from the ve terminal ve terminal (high potential) (low potential) through the resistor (load). So, the direction of the current is from high potential to low potential. Note: from Ohm s Law: = V R = 9 0.47 =19.1A and P = 2 R = 172W. The resistor gets very hot!! When the switch is closed, the batteries cause charges to flow through the LOAD (a bulb) and around the circuit. When the switch is opened, the charges can no longer flow around the circuit. R A generic simple circuit: Note: current does not get used up in a circuit. t is the same all the way round even in the LOAD!
Consider a battery not in a ΔV = circuit (i.e., not delivering a Mechanical analog of a simple circuit... current). f the potential at a is a b potential difference is V a and the potential at b is V b (note V a > V b ) then the b a c V a V b =, d and is called the electromotive force (emf) of the battery. c a R b d Consider a battery in a circuit. Typically, the connecting wires ac and bd have negligible resistance compared with the load (R), so: ΔV ac (= R ac ) 0 and ΔV db (= R db ) 0. V a = V c and V d = V b so the full emf is across the load. xample: Consider a flashlight. The resistance of a bulb is ~ 5Ω; the resistance of the connections between the a b c d The source of emf (i.e., the battery) is like an elevator doing work to lift charge from a position of low potential ( ve), V b, to a position of high potential ( ve), V a, where: = V a V b. n a mechanical analog, marbles (charges) are raised to a point of higher potential energy V c at c. They fall from the top of the track, to a position of lower potential V d at d, losing potential energy. At the bottom, work is done to lift them back to the point of high potential. bulb and the battery is <10 3 Ω.
Mechanical analog of a battery: V a V = V a V b δq V b ncreasing potential energy A battery acts like a charge elevator doing work to move charges from the ve to the ve terminal and raising their potential energy. The potential difference ( V) is a fixed quantity so the battery raises each charge ( δq) the same amount of potential energy ( δu = δq.v = δw). The total power Question 25.2: Find the power dissipated in a resistor connected across a constant potential difference of 120V if its resistance is 5Ω. required is: P = δw δt = δ δt ( nδqv) = V work done by the battery The amount of current supplied, i.e., the rate of flow of charge depends on the rate the elevator moves; faster for more current, slower for less current. The height doesn t change, so V remains constant even though the current changes.
5Ω = 120 V 5Ω Since the emf of the battery is applied across the load resistor, the current is = V R = R = 120 V 5Ω = 24.0 A. The power dissipated by the 5Ω resistor is: P = 2 R = (24.0 A) 2 5Ω = 2880 W 2.88 kw. What is the power dissipated by the battery? P = (120 V) (24.0 A) = 2.88 kw. Also, since the full emf of the battery is across the resistor: P = 2 R = (120 V)2 5Ω = 2.88 kw. = 120 V Let s look at the potential around the circuit: We ll start at c and move clockwise V da = V bc = Note that the current is constant around the circuit: = R = 120 V = 24.0 A. 5 Ω
δq Question 25.3: When an electric current passes through a load, the charges lose energy producing thermal energy in the load. Do the charges that make up the current lose (a) kinetic energy, (b) potential energy, or (c) a combination of the two? Charges move from a point of high potential () to low potential ( ), so their potential energy changes ( δu = δq.v). But the kinetic energy of the charges does not change! f it did, it would mean that their speed changes. f the speed decreased the charges would build up in some region... if the speed increased the charges would be s t r e t c h e d out! Neither of these things happen because the rate of flow of charge in a circuit (i.e., the current) is always the same everywhere... therefore, there can be no change in speed. So, the answer is (b), charges lose potential energy only.
QUSTON: what happens when we connect different loads (resistors) together? There are two ways they can be connected. xample of resistors N SRS. Resistances N SRS: R 1 R 2 R 3 V R 1 R 2 R 3 R eq R 1 R 2 R 3 xamples of resistors N PARALLL. R 1 R 1 V equivalent resistor The current through each resistor, R 1, R 2 and R 3, is the same. An equivalent resistor will produce the same potential difference (V) for the same current (): R 2 R 3 R 3 R 2 i.e., V = R eq = (R 1 R 2 R 3 ) R eq = R 1 R 2 R 3. n general with resistors in series: R eq = i R i.
Resistors in series... X 12 V 4Ω 2Ω Unfortunately, real batteries come with internal 2 1 resistance... as in this example Total resistance R 1 R 2 = 2Ω 4Ω = 6Ω. = = 12 V = 2.0 A. R total 6Ω Potential difference across R 1 V 1 = R 1 = 4 volts. Potential difference across R 2 V 2 = R 2 = 8 volts. Note: V 1 V 2 = 12 volts =. Look at the potential difference around the circuit: V 12 V V 1 = 4 V V 2 = 8 V Question 25.4: An 11.0Ω resistor is connected across a battery with an emf of 6.00 V and internal resistance 1.00Ω. Find: (a) the current in the circuit, (b) the voltage across the terminals of the battery, (c) the power supplied by the battery, (d) the power delivered to the external resistor, and (e) the power delivered to the battery s internal resistance. X
Note the resistors are in series... 11Ω = 6 V, r = 1Ω (a) The total resistance in this circuit is: R = 11Ω 1Ω = 12Ω. = R = 6 V 12Ω = 0.5 A. (With no internal resistance R = 11Ω and 0.55 A.) (b) The potential difference across battery terminals: V = r = (6 V) (0.5 A 1Ω) = 5.5 V. Note that this is less than the emf... the emf of a battery is the potential difference across the battery terminals with no load, i.e., when it not supplying current. Question 25.5: A battery has an emf equal to 12.0 V and an internal resistance of 4.00 Ω. What value of external resistance R should be placed across the terminals of the battery to obtain maximum power delivered to the resistor? (c) P = = (6 V) (0.5 A) = 3.0 W. (d) P R = 2 R = (0.5 A) 2 11Ω = 2.75 W. (e) P r = 2 r = (0.5 A) 2 1Ω = 0.25 W.
R Resistors N PARALLL: R 1 1 2 R 2 R eq The power delivered to the resistor is P = 2 R, where = = 12.0 V r = 4.00Ω (R r), i.e., P = 2 R (R r) 2. For maximum power, dp dr = 0. dp dr = (R r)2 2 2 2 R(R r) (R r) 4 = (R r)2 2 2 R (r R)2 (R r) 3 = (R r) 3. Hence, dp = 0 when R = r = 4.00Ω. This is the dr condition for maximum power. Also true in ac circuits. 3 R 3 V Note: the potential difference across each resistor ( V i = i R i ) is the same but if R 1 R 2 R 3 the current through each resistor will be different. However, = 1 2 3. (This is Kirchoff s 1st rule... later.) = V V V = V = V 1 1 1, R 1 R 2 R 3 R eq R 1 R 2 R 3 i.e., 1 R eq = 1 R 1 1 R 2 1 R 3. n general, with resistors in parallel, 1 1 = i. R eq R i V equivalent resistor
a (a) b b (b) (b) (a) Question 25.6: f the bulbs and batteries are identical, in which circuit are the bulbs brighter or are they both the same? The brightness of a bulb is determined by the power dissipated; the more power, the brighter the bulb. n (a) the current through each bulb is a = 2R. Therefore, the power dissipated by each bulb in (a) is: ( ) 2 R = 2 4R. P a = 2R n (b) since the potential difference across the bulbs is the same, the current through each bulb is b = R. Therefore, the power dissipated by each bulb in (b) is: P b = R ( ) 2 R = 2 R. P b = 4P a, so the bulbs are brighter in (b).
a b c Let the resistance of each bulb be r. f is the emf of each battery, the power dissipated in each circuit is 2 R tot, Question 25.7: f the bulbs and batteries are identical, which circuit draws most power from the battery? where R tot is the total equivalent resistance. n (a): R tot = r. a b c Assume the batteries have zero internal resistance. n (b): n (c): 1 1 = R tot (r r) 1 (r r) = 1 2r 1 2r = 1 r. i.e., R tot = r. 1 R tot = 1 3r 1 3r 1 3r = 3 3r = 1 r. i.e., R tot = r. Since the equivalent resistances of (a), (b) and (c) are the same, the power drawn from the batteries (and the currents) is the same in each case!
3Ω 3 1 2 2Ω 2Ω 4 4Ω Question 25.8: f the battery in the circuit shown has negligible internal resistance, find (a) the current in each resistor, and (b) the power delivered by the battery. 3Ω 3 (a) First, we calculate 1, but to do that we need the total resistance of the circuit. We recognize that the right hand group of 3 resistors are all in parallel, so the equivalent resistance is given by: 1 = 1 R eq 2Ω 1 2Ω 1 4Ω = 2 2 1 4Ω = 5 4Ω. 1 2 2Ω 2Ω 4 4Ω R eq = 4 5Ω = 0.8Ω. So, here is the equivalent circuit: 3Ω 1 6 V 0.8Ω The 3Ω resistor is in series with the equivalent resistance of 0.8Ω, so 6 V 1 = = 1.58 A. ( 3Ω 0.8Ω)
To determine the currents 2, 3 and 4 through the parallel combination we need to find the potential difference across them, i.e., V ab. By Ohm s Law: 3Ω 1.58A a 6 V 0.8Ω b V ab = 1.58 A 0.8Ω = 1.26 V. a 3 2 2Ω 1.26 V 2Ω 4Ω b 4 The potential difference across each of the three resistors (in parallel) is the same: 2 = 1.26 V 2Ω = 0.63 A: 3 = 1.26 V 4Ω = 0.32 A: 4 = 1.26 V 2Ω = 0.63 A. (Note that: 1 = 2 3 4 = 1.58 A, which it must since current cannot be lost or gained.) (b) The power delivered by the battery: P = 1, where is the emf of the battery. P = 6 V 1.58 A = 9.48 W. Also, P = 2 R tot, where R tot is the equivalent resistance of the ensemble of resistors. From (a) R tot = 3Ω 0.8Ω = 3.8Ω. P = (Rounding errror.) (6 V)2 3.8Ω = 9.47 W. You could also calculate the power dissipated by each of the four resistors, i.e., P = 2 i i R i but it takes more time!
c 4Ω 4Ω a b 4Ω 4Ω d Question 25.9: Four 4Ω resistors are arranged as shown. (a) a c 4Ω 4Ω 4Ω 4Ω d b a 4Ω c 4Ω 4Ω d 4Ω b (a) What is the equivalent resistance between a and b? (b) What is the equivalent resistance between a and b if an additional 4Ω resistor was connected across c and d? The circuit is equivalent to two 8Ω resistors in parallel. 1 = 1 R eq 8Ω 1 8Ω = 1 4Ω, a i.e., R eq = R ab = 4Ω. 8Ω 8Ω b
(b) c f we put a battery across 4Ω 4Ω ab, equal currents will a b flow through c as through 4Ω 4Ω d because the two routes d are identical (a resistance 4Ω c 4Ω of 8Ω). Since the resistors a b are all the same ( 4Ω), the potential at c is the same as 4Ω d 4Ω the potential at d! Therefore, there s no potential difference between c and d, so no current will flow through any resistor connected across cd. As no extra current is required from the battery, the equivalent resistance remains the same! With more complex circuits, e.g., with more than one source of emf, we must follow certain rules (which we ve already come across)... Kirchhoff s Rules: [1] The algebraic sum of currents at a junction is zero, i.e., when currents meet at a junction 2 1 3 5 i i = 0. 4 Don t always look for a formula!! 1 2 3 4 5 = 0, i.e., 1 2 3 = 4 5.
[2] The algebraic sum of the potential differences around a circuit is zero, i.e., around a circuit i V i = 0. xample... Remember V a > V b and V c > V d, i.e., the potential difference across a resistor decreases in the direction of the current, and across the battery V f > V e. V f f e V ab V cd a b c d a b V ab c V cd 12V X 8 V? ΔV Y 6 V Question 25.10: (a) What is the potential difference ΔV across the unknown circuit element in the direction of the current? (b) Which point, X or Y, is at the higher potential? d e d V ab V cd = 0 (Kirchoff s 2nd rule), i.e., V ab V cd =.
A 12V X 8 V? ΔV Y 6 V (a) By Kirchoff s 2nd rule, in the direction of the current, and starting from A, we have: 12 V ΔV 6V 8 V = 0 ΔV = 2 volts. (b) Since ΔV > 0 the potential increases in the direction of the current so the potential at Y is higher than at X. So it cannot be a resistor; it is a source of emf, like a 2V battery. Y V ΔV 6 V X Question 25.11: n the circuit shown below, the batteries have negligible internal resistance. Find (a) the current in each branch of the circuit, the potential difference between point X and Y, and (c) the power supplied by each battery. 4Ω X 3Ω 12 V 6Ω 12 V Y 12 V 8 V A
1 4Ω 1 X 2 3 3Ω 12 V 6Ω 12 V A B 3 1 Y 2 (a) Choose the directions of the currents arbitrarily. Using Kirchoff s rules we have: 1 = 2 3............ (1) n circuit A going clockwise (starting at battery): 2 1 4Ω 1 X 2 3 3Ω 12 V 6Ω 12 V A B 1 Y 2 n circuit B we have (starting at X): 3 2 12 V 6 3 12 V 3 2 12 V 4 1 12 V 6 3 12 4 1 6 3 = 0, i.e., 4 1 6 3 = 12......... (2) 3 2 12 6 3 = 0 i.e., 3 2 6 3 = 12......... (3) Using (1), substitute for 1 in (2). Then (2) becomes: 4( 2 3 ) 6 3 =12 i.e., 4 2 10 3 = 12............ (4) quations (3) and (4) are a pair of simultaneous linear equations.
1 4Ω 1 X 2 3 3Ω 12 V 6Ω 12 V A B 3 1 Y 2 Solving equations (3) and (4) as a pair of simultaneous linear equations gives: 2 = 0.90 A and 3 = 1.56 A. (The negative sign mans we chose the wrong direction for 2!) 1 = ( 0.90 A) 1.56 A = 0.66 A. (b) V XY = 3 6Ω = 1.56 A 6Ω = 9.36 V. (c) P =. P A = 12 V 0.66 A = 7.92 W, and P B = 12V 0.90 A = 10.80 W. 2 Simple RC circuit: Charging the capacitor: Close S 1 and leave S 2 open at t = 0. Charges flow through resistor R and onto the capacitor C. At some time t, let the charge on the capacitor be q. Using Kirchoff s 2nd rule: V R V C = 0, i.e., i 1 R q C = 0, or dq dt q RC R = 0. This is a 1storder differential equation that tells us how q varies with t. R i 1 R i 1 C V R S 2 S 1 V C S 2 C S 1 nitial conditions
The solution is q(t) = C(1 e t RC ) = Q F (1 e t τ ) where Q F (= C) is the final (maximum) charge and the quantity τ ( = RC) is called the TM CONSTANT, which gives us a measure of how quickly the capacitor becomes charged. q(t) Q F asymptote 0.632Q F Since q(t) = C(1 e t RC ), the current i 1 at any time t is: i 1 (t) = dq dt = RC e t R =! e t RC =! e t τ, where! (= R ) is the initial current in the circuit the instant S 1 is closed.! = R 0.368! i 1 (t) When t = τ, i i 1 =! e 1 1 (τ) = 0.368!. Also V R (t) = i 1 (t)r. V R (τ) = 0.368. When t = τ = RC τ = RC q(τ) = Q F 1 e 1 Also, since V C (t) = q(t) C then V C (τ) = 0.632Q F C ( ) = 0.632Q F. and Q F = C, = 0.632. t τ = RC t Note: when connected across a battery, an uncharged capacitor acts like an shortcircuit so the current is a maximum ( i 1 (0) max =! ). a fully charged capacitor acts like an opencircuit, i.e., an infinite resistance, so the current is zero ( i 1( ) min = 0).
Discharging the capacitor: Now, start with a charge Q! on the capacitor and switch S 1 open; so the capacitor is charged. i 2 R Q! i 2 Close S 2 at t = 0. The capacitor discharges so charges (i.e., a current i 2 ) flows through R. Then at some time t, we have from Kirchoff s 2nd rule, in the direction of the arrow (and current i 2 (t)*): V C V R = 0, i.e., i 2 (t)r q C = 0, or S 2 C S 1 dq dt q RC = 0. * Note that i 2 during discharging is in the opposite direction to i 1 during charging. The solution is: 0.368Q! When t = τ: q(t) t RC q(t) = Q! e Q! τ = RC = Q! e t τ. q(τ) = Q! e t τ = Q! e 1 = 0.368Q! Also, since V C (t) = q(t) C, then V C (τ) = 0.368Q!. C t
Since q(t) = Q! e t RC, the current at any time t is: i 2 (t) = dq dt = Q! RC e t RC =! e t τ, where! = Q! RC, i.e., the initial current at t = 0 when S 2 is closed. i 2 (t) During the charging process we have... V R V C = 0 i.e., V R V C =. V(t) V C τ = RC t V R t 0.368! τ = RC During the discharging process we have...! When t = τ: V(t) V C V R = 0 i.e., V R = V C. i 2 (τ) =! e t τ = Q! RC e 1 = 0.368 Q! RC. Also, since V R (t) = i 2 (t)r, then V R (τ) = 0.368Q! C = V C (τ). τ = RC V R,V C t
Question 25.12: n the circuit shown below, switch S is opened after having been closed for a long time so the capacitor is fully discharged. Four seconds after S is opened, the potential difference across the resistor is 20.0 V. What is the resistance of the resistor? i(t) 50 V R 2µF S i(t) 50 V R With switch S closed, the capacitor is completely uncharged and the potential difference across R is 50 V. When switch S is opened at t = 0, the capacitor begins to charge and the potential difference across R decreases. When t = 4s, V R = 20 V, what is the value of R? We know: V R V R (t) = i(t)r =! Re t RC. But initially: V R (0) =! R =. V R (t) = e t RC, i.e., e t RC = V R(t). Take natural logarithms of both sides... t RC = ln V R(t) ( ). ( ) = 0.916 4s (2 10 6 F) R = ln 20 V 50 V i.e., R = 2µF S 4s (2 10 6 F) 0.916 = 2.18 106 Ω. The time constant is τ = RC = 4.37s.
nergy in a RC circuit: During charging, a current flows through the battery and resistor R. The instantaneous power dissipated by the battery is i(t). Total energy dissipated by the battery to fully charge the capacitor is i(t).dt = dq.dt = dq = Q F. dt 0 i(t) R 0 But, when the capacitor is charged to Q F, the energy stored in the capacitor is 1 2 Q F. So, using conservation Q F of energy, the energy dissipated by the resistor is: C Q F 1 2 Q F 1 2 Q F. 0 i(t) R So, during charging, onehalf of the energy dissipated by the battery (= Q F ) is stored in the capacitor = 1 2 Q F and the other half is dissipated by the resistor = 1 2 Q F. Note, if the instantaneous current is i(t), the instataneous power dissipated by R is: P R (t) = i 2 (t)r, and the instantaneous power output of the battery is i(t). So, the energy stored in the capacitor at time t is C U C (t) = i(t) i 2 (t)r. work done by battery energy stored in capacitor