Upcoming SlideShare
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Standard text messaging rates apply

# Ch27 circuits

97

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total Views
97
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
4
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Transcript

• 1. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Ch 27: Circuits p705 - 723 (exclude 27-8:meters) 2 lectures An emf device is one which maintains a potential difference between a pair of terminals. It provides an emf, , => does work on the charge carriers. It must have a source of energy. emf = electromotive force (an outdated phrase). examples battery, electric generator, fuel cell thermopiles, solar cells, living systems (eels, people) Ch27-1/32 Work, energy, emf Represent the emf by an arrow from -ve to +ve terminal (i.e. from lower to higher potential). Circle on tail to distinguish from arrow for current direction. Current is same all around a circuit. Define emf of an emf device as amount of work done moving a charge dq from low to high potential terminal. An ideal emf device has no internal resistance so V = Real emf devices have internal resistance => so V < when current is flowing Ch27-2/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 1
• 2. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Calculating current in a single-loop circuit Energy Method P=i2R tells us the energy dissipated in resistor in time dt is i2Rdt In same time dt, charge dq = i dt will have moved thru battery B which, from defn of emf, will do work Conservation of energy => work of battery = heat dissipated: where emf is energy per unit charge transferred to moving charges by battery. Hence: (for an ideal battery) Ch27-3/32 Calculating current in a single-loop circuit Potential Method Start at any point in circuit - go around in either direction, - add up potential differences. After a loop, the change in potential must be 0. Kirchhoff’s loop rule In general, the algebraic sum of changes in potential encountered in a complete traversal of any loop in a circuit must be zero. Start at point a where potential is Va. go clockwise. Potential difference across ideal battery is + . Potential drops across R (high to low potential) , i.e. V = -iR. Note if we went other way we get same result. Ch27-4/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 2
• 3. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Calculating current in a single-loop circuit Potential Method (cont) For a move thru a resistance, R, in direction of current, change in potential is -iR; V = iR in the opposite direction it is +iR For a move thru an ideal emf device in direction of emf arrow, change in potential is + ; V= in opposite direction it is - . Resistance case depends on current direction: emf case depends on direction of emf & current Ch27-5/32 Question The current in a single loop circuit is shown, with a battery and a resistance. (a) should the emf arrow be right or left? At points a, b, c rank, largest first (b) the magnitude of the current (c) the electric potential Ch27-6/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 3
• 4. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question The current in a single loop circuit is shown, with a battery and a resistance. Ans: right since i flows from high to low potential (a) should the emf arrow be right or left? and emf arrow goes from low to high potential At points a, b, c rank, largest first (b) the magnitude of the current Ans (b): Same everywhere (c) the electric potential Ans (c): high at b, same at a & c since change is -iR crossing R Ch27-7/32 Internal resistance in a real battery Consider internal resistance which is part of any real battery. Order in circuit of ideal emf vs resistance is not important. In practice resistance is throughout battery. Apply loop rule going clockwise from a. For r = 0 this reduces to previous eqn i = /R. Potential across battery is i.e. greater current => lower effective voltage Ch27-8/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 4
• 5. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Potential across a real battery So the internal resistance reduces the effective potential difference by 1/(1+r/R). Ch27-9/32 Internal resistance in a real battery Right is a graphical representation of potential at various points in circuit, starting at a and moving clockwise. The shape inside emf device is not particularly realistic, but is representative. Ch27-10/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 5
• 6. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Resistance in series Current must be same in all resistances since it is same everywhere in a closed circuit. Sum of potential differences across resistances is equal to applied potential difference V Can replace resistances in series by an equivalent resistance Req with same potential difference V and current, i. Ch27-11/32 Resistance in series To derive an expression for Req apply loop rule, starting at a, clockwise. In circuit (b) and since currents are same which generalizes to Ch27-12/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 6
• 7. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question If R1 > R2 > R3, rank (greatest first) the 3 resistances in terms of (a) the current flowing thru them (b) the potential difference across them. Ch27-13/32 Question If R1 > R2 > R3, rank (greatest first) the 3 resistances in terms of (a) the current flowing thru them (b) the potential difference across them. Ans(a): the current is the same everywhere in circuit Ans(b): V = iR, so order is as with resistances Ch27-14/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 7
• 8. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Grounding a circuit The symbol means circuit is connected to ground at that point. The earth is considered to be a giant conductor and therefore at the same potential everywhere. Define potential at that point to be zero. Ch27-15/32 Sample problem (a) What is current, i, in circuit? Key idea: Use loop rule going counterclockwise (arbitrary) from `a’. Since emf1 > emf2, current flows clockwise as shown (but this is not critical) For a move thru a resistance, R, in For a move thru an ideal emf device the direction of current, the in the direction of the emf arrow, change in potential is -iR; in the the change in potential is + emf ; opposite direction it is +iR in the opposite direction it is - emf Note we get same eqn going clockwise. Solve for i to get i = 240 mA. Ch27-16/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 8
• 9. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem (cont) (b) What is the potential difference between terminals of B1 Key idea: Sum the potential differences from clockwise from b to a: Get same answer going counterclockwise from b to a, but there are 3 elements to consider => more effort to get solution Ch27-17/32 Question A battery has an emf of 12 V and an internal resistance of 2Ω. Is the terminal to terminal potential difference >, =, or < 12 V if the current in the battery is (a) from the -ve to +ve terminal (b) from the +ve to -ve terminal (c) 0 Ch27-18/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 9
• 10. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Question A battery has an emf of 12 V and an internal resistance of 2Ω. Is the terminal to terminal potential difference >, =, or < 12 V if the current in the battery is (a) from the -ve to +ve terminal Ans: move in direction of current, ∆V= emf - iR < 12V (b) from the +ve to -ve terminal Ans: move in direction of current, ∆V=-emf - iR >12V (c) 0 Ans: no drop over resistance => 12 V For a move thru a resistance, R, in For a move thru an ideal emf device the direction of current, the in the direction of the emf arrow, change in potential is -iR; in the the change in potential is + emf ; opposite direction it is +iR in the opposite direction it is - emf Ch27-19/32 Multiloop Circuits What are the 3 currents in figure? There are 2 junctions, b and d and three branches with currents: bad; bcd; and bd. Directions are arbitrary. At junction d: i1 + i3 = i2 . Follows from charge conservation. One gets the same at b. (Kirchhoff’s) Junction Rule: sum of currents entering a junction = sum of currents leaving it. Now apply loop rule: Three loops: left, right and combined (outer). Only need 2 more eqns to solve for 3 unknowns. Ch27-20/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 10
• 11. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Multiloop Circuits (cont) Left-loop, counterclockwise: Right-hand loop, counterclockwise Along with previous equation, i1 + i3 = i2 => 3 eqns in 3 unknowns => can be solved. If we had used the big outer loop we would have: but this adds no new info since it is sum of above 2 eqns Ch27-21/32 Resistances in parallel In parallel and in series have same meaning as with capacitors. When a potential difference V is applied across resistances connected in parallel, resistances all have same potential difference V. Resistances connected in parallel can be replaced with an equivalent resistance Req that has same potential difference V and same total current i as actual resistances. Ch27-22/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 11
• 12. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Resistances in parallel (cont) What is Req. for resistances in parallel? x x where V is potential difference between a and b. y Apply junction rule at x and y to get currents i2+i3 shown. At junction near point a: => hence: which generalizes to: Ch27-23/32 Resistances in parallel (cont) For resistors in parallel, the equivalent resistance is smaller than any of the combined resistances (there are more places for current to go). For 2 resistors in parallel we get Remember which way is up by looking at dimensions. Ch27-24/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 12
• 13. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Summary of parallel and series resistors parallel series -same potential difference across all same current in all -currents add up capacitors parallel series -same potential difference across all same charge on all -charges add up Ch27-25/32 RC circuits Study circuits where current varies with time. RC series circuit has ideal battery, resistance R and initially uncharged capacitor C (q(0) = 0). Switch to a => current flows to capacitor. At any given time, potential on capacitor is VC = q/C (q=CV). When VC = emf => no more current flow & Find q(t) and i(t) in the circuit. The loop rule can be applied at any point in time. Go clockwise from the -ve terminal arrow same dir current same dir potential drop across C Ch27-26/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 13
• 14. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons RC circuit: charging capacitor RC circuit eqn solution of RC circuit eqn First look at the limits at t = 0 q(0) = 0, as required. at very large t, the equilibrium charge becomes, Ch27-27/32 RC circuits: charging capacitor (cont) Theorem: The expression is a solution of the RC circuit eqn Proof: QED Text outlines a strategy for solving this type of differential eqn. but given the solution, the above is a rigorous proof it is a solution. Ch27-28/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 14
• 15. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons RC circuits: charging capacitor (cont) What is current flow as a function of time while charging an RC circuit? So at t = 0 as if no C and at large t, i(t) = 0 i.e.: at start, capacitor is a conducting wire and at end it is like an open circuit (no connection). What is voltage on C with time? V(0) = 0 and V(large t) = Ch27-29/32 RC circuits: charging capacitor (cont) Rewrite using time constant Units: =>t/τ = t/RC is dimensionless Any exponential must be dimensionless. What is charge after 1 time constant? i.e. 63% of charge in first time constant (86% in 2τ etc) R = 2kΩ, C = 1µF, emf = 10V τ = 2ms Ch27-30/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 15
• 16. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons RC circuits: discharging capacitor Toggle the switch to b. => discharging capacitor with no emf in circuit. Loop eqn gives, as before, but no emf Theorem: is a solution of above differential eqn. if initial charge on capacitor is qo Proof: (do substitution as before 27-25) Note: q(0) = qo and q(large t) = 0 After τ, charge reduced to 37% of initial value. Ch27-31/32 RC circuits: discharging capacitor (cont) The current also decays exponentially with time. Initial value is qo/RC. Note: ignore the - sign in eqn. It just tells us it is decaying/decreasing. Ch27-32/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 16
• 17. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons More complex RC circuits Consider the circuit shown. Theorem: We can solve circuit R1 by replacing the capacitors and resistors by their equivalent C1 C2 single elements and using previous eqn. Proof: Use loop rule as before R2 This is same eqn as before but using equivalent R and C => solution follows as before. QED Ch27-33/32 Exponential decay in general The above equation for the discharging capacitor can be rewritten as An equivalent equation and its solution are Applies to any situation where rate of change in a quantity is proportional to the quantity. Major example, radioactive decay. Ch27-34/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 17
• 18. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Ch 27: Circuits p705 - 723 (exclude 27-8:meters) 2 lectures These are some problems from the end of this chapter which were not covered in class Ch27-35/32 Sample problem RC circuits The text frames a problem in terms of the charge which builds up to 30 kV on a car when racing. When it stops in the race pit, the car and the earth are considered as a capacitor (C = 500 pF) with 4 tires providing a resistance to ground (Rtire = 100 GΩ). As long as there is more than 50 mJ of energy stored in the capacitor, a spark can cause a fire. How long must they wait before the stored energy leaks away? Making use of earlier theorem about complex RC circuits Ch27-36/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 18
• 19. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem RC circuits Stored electric potential energy Charge at t Initial charge (given C and Vo) Solve for time t at which stored electric potential energy is any value we care about, here 50 mJ Ch27-37/32 Sample problem RC circuits Time for discharge to below 50 mJ is 9.4 s. Would they wait? Not in a race, so one option is to make tires more conductive so R = 10 GΩ instead of 100 => 0.94 s Any other suggested solutions? Ans: Grounding cable Ch27-38/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 19
• 20. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem (circuits) p716/717 Given the circuit on the left with an ideal battery, (a) find the current thru the battery. Key ideas: assign currents as shown in (b) where we recognize that the current thru the battery and R1, R4 are the same. Our goal is to get resistances in series since they have a common current => replace R2 and R3 with R23 as in (c). R2 and R3 are in parallel so: Ch27-39/32 Sample problem (cont) Now apply the loop rule to (c) which can easily be solved to give i1. (b) What is the current thru R2? Key idea: Given i1 we can determine V23 and from this i2 Ch27-40/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 20
• 21. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Sample problem (cont) (c) What is the current thru R3? Key idea: We could do this the same as part (b) or apply the junction rule at point b in fig (b). Ch27-41/32 Another sample problem p717 All batteries are ideal batteries. What is the current in each branch? Given emfs shown, it seems reasonable to assign current directions as shown, but these are arbitrary for now. Ch27-42/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 21
• 22. Rogers: Lectures based on Halliday, Resnick and Walker’sFundamentals of Physics, Copyright 2005 by Wiley and Sons Another sample problem (cont) Key ideas: Apply junction and loop rules Junction rule at a or b gives Loop rule for left hand loop, going counterclockwise from a Loop rule for right hand loop, going clockwise from a 3 eqns, 3 unknowns, easy to solve -ve sign => current in opposite direction to that shown. Ch27-43/32 Sample problem (27-4 p718) A neat example of an electric eel. Read it! Note useful simplification of circuit since all points b are at the same potential in left circuit => right circuit is equivalent => can apply rule regarding resistances in parallel (same potential across all). Ch27-44/32Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to CopyrightProtection. 22