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Overview Discuss Test 1 Review –Kirchoff's Rules for Circuits –Resistors in Series & Parallel RC Circuits Text Reference: Chapter 27,

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UNIT: Ampere = A = C/s If there is a potential difference between two points then, if there is a conducting path, free charge will flow from the higher to the lower potential. The amount of charge which flows per unit time is defined as the current I, i.e. current is charge flow per unit time. Current- a Definition i.e. no longer electrostatics.

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Resistors: Purpose is to limit current drawn in a circuit. Resistors are basically bad conductors. Actually all conductors have some resistance to the flow of charge. Devices Resistance Resistance is defined to be the ratio of the applied voltage to the current passing through. V I I R UNIT: OHM =

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Resistivity L A E I Property of bulk matter related to resistance : The flow of charge is easier with a larger cross sectional area, it is harder if L is large. eg, for a copper wire, ~ -m, 1mm radius, 1 m long, then R .01 The constant of proportionality is called the resistivity The resistivity depends on the details of the atomic structure which makes up the resistor (see chapter 27 in text)

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Ohm's Law Demo: Vary applied voltage V. Measure current I Does ratio (V/I) remain constant?? V I slope = R V I I R Only true for ideal resisitor!

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Lecture 11, CQ 1 Two cylindrical resistors, R 1 and R 2, are made of identical material. R 2 has twice the length of R 1 but half the radius of R 1. –These resistors are then connected to a battery V as shown: V I1I1 I2I2 –What is the relation between I 1, the current flowing in R 1, and I 2, the current flowing in R 2 ? (a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2 The resistivity of both resistors is the same . Therefore the resistances are related as: The resistors have the same voltage across them; therefore

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Kirchoff's First Rule "Loop Rule" or “Kirchoff’s Voltage Law (KVL)” "When any closed circuit loop is traversed, the algebraic sum of the changes in potential must equal zero." This is just a restatement of what you already know: that the potential difference is independent of path! KVL: We will follow the convention that voltage drops enter with a + sign and voltage gains enter with a - sign in this equation. RULES OF THE ROAD: R1R1 R2R2 I Move clockwise around circuit: R1R1 R2R2 I IR 1 IR 2 0 0

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Kirchoff's Second Rule "Junction Rule" or “Kirchoff’s Current Law (KCL)” In deriving the formula for the equivalent resistance of 2 resistors in parallel, we applied Kirchoff's Second Rule (the junction rule). "At any junction point in a circuit where the current can divide (also called a node), the sum of the currents into the node must equal the sum of the currents out of the node." This is just a statement of the conservation of charge at any given node.

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Resistors in Series The Voltage “drops”: Whenever devices are in SERIES, the current is the same through both ! This reduces the circuit to: a c R effective Hence: a b c R1R1 R2R2 I

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Resistors in Parallel What to do? But current through R 1 is not I ! Call it I 1. Similarly, R 2 I 2. How is I related to I 1 & I 2 ?? Current is conserved! a d a d I I I I R1R1 R2R2 I1I1 I2I2 R V V KVL Very generally, devices in parallel have the same voltage drop

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Loop Demo a d b e c f R1R1 I R2R2 R3R3 R4R4 I KVL:

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Lecture 11, CQ 2 Consider the circuit shown. –The switch is initially open and the current flowing through the bottom resistor is I 0. –Just after the switch is closed, the current flowing through the bottom resistor is I 1. –What is the relation between I 0 and I 1 ? (a) I 1 < I 0 (b) I 1 = I 0 (c) I 1 > I 0 R 12 V R I

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Lecture 11, CQ 2 Consider the circuit shown. –The switch is initially open and the current flowing through the bottom resistor is I 0. –After the switch is closed, the current flowing through the bottom resistor is I 1. –What is the relation between I 0 and I 1 ? (a) I 1 < I 0 (b) I 1 = I 0 (c) I 1 > I 0 Write a loop law for original loop: R 12 V R I a b -12V + I 1 R = 0 I 1 = 12V/R Write a loop law for the new loop: -12V -12V + I 0 R + I 0 R = 0 I 0 = 12V/R

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…or, Lecture 11, CQ 2 Consider the circuit shown. –The switch is initially open and the current flowing through the bottom resistor is I 0. –After the switch is closed, the current flowing through the bottom resistor is I 1. –What is the relation between I 0 and I 1 ? (a) I 1 < I 0 (b) I 1 = I 0 (c) I 1 > I 0 The key here is to determine the potential (V a -V b ) before the switch is closed. From symmetry, (V a -V b ) = +12V. Therefore, when the switch is closed, NO additional current will flow! Therefore, the current after the switch is closed is equal to the current after the switch is closed. R 12 V R I a b

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Junction Demo Outside loop: Top loop:Junction: I1I1 R R R I2I2 I3I3

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a b R C I I t q RC 2RC 0 CC

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Overview of Lecture RC Circuit: Charging of capacitor through a Resistor RC Circuit: Discharging of capacitor through a Resistor Text Reference: Chapter 27.4, 28.2, 28.6

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RC Circuits R C I I Add a Capacitor to a simple circuit with a resistor Recall voltage “drop” on C? Upon closing circuit Loop rule gives: Recall that Substituting: Differential Equation for q!

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Compare with simple resistance circuit Simple resistance circuit: –Main Feature: Currents are attained instantaneously and do not vary with time!! Circuit with a capacitor: –KVL yields a differential equation with a term proportional to q and a term proportional to I = dq/dt. Physically, what’s happening is that the final charge cannot be placed on a capacitor instantly. –Initially, the voltage drop across an uncharged capacitor = 0 because the charge on it is zero ! –As current starts to flow, charge builds up on the capacitor, the voltage drop is proportional to this charge and increases; it then becomes more difficult to add more charge so the current slows

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The differential equation is easy to solve if we re-write in the form: or equivalently, Integrating both sides we obtain, Exponentiating both sides we obtain, If there is no initial charge on C then: Thus, Integration constant We have to find q such thatis satisfied. Determines integration constant

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Charging the Capacitor Charge on C Max = C 63% Max at t=RC t q 0 cc RC 2RC I 0 t R Current Max = /R 37% Max at t=RC

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Lecture 12, CQ 1 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –What is the value of the current I 0+ just after the switch is thrown? (a) I 0+ = 0 (b) I 0+ = /2R (c) I 0+ = 2 /R (a) I = 0 (b) I = /2R (c) I > 2 /R 1B – What is the value of the current I after a very long time? 1A

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Lecture 12, CQ 1 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –What is the value of the current I 0+ just after the switch is thrown? (a) I 0+ = 0 (b) I 0+ = /2R (c) I 0+ = 2 /R 1A Just after the switch is thrown, the capacitor still has no charge, therefore the voltage drop across the capacitor = 0! Applying KVL to the loop at t=0+, I R I R - = 0 I = /2R

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Lecture 12, CQ 1 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –What is the value of the current I 0+ just after the switch is thrown? (a) I 0+ = 0 (b) I 0+ = /2R (c) I 0+ = 2 /R 1A 1B (a) I = 0 (b) I = /2R (c) I > 2 /R – What is the value of the current I after a very long time? The key here is to realize that as the current continues to flow, the charge on the capacitor continues to grow. As the charge on the capacitor continues to grow, the voltage across the capacitor will increase. The voltage across the capacitor is limited to ; the current goes to 0.

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Lecture 12, CQ 2 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –At time t=t 1 = , the charge Q 1 on the capacitor is (1-1/e) of its asymptotic charge Q f =C . –What is the relation between Q 1 and Q 2, the charge on the capacitor at time t=t 2 =2 ? (a) Q 2 < 2 Q 1 (b) Q 2 = 2 Q 1 (c) Q 2 > 2 Q 1 Hint: think graphically!

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(a) Q 2 < 2 Q 1 (b) Q 2 = 2 Q 1 (c) Q 2 > 2 Q 1 From the graph at the right, it is clear that the charge increase is not as fast as linear. In fact the rate of increase is just proportional to the current (dq/dt) which decreases with time. Therefore, Q 2 < 2Q 1. The charge q on the capacitor increases with time as: So the question is: how does this charge increase differ from a linear increase? At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –At time t=t 1 = , the charge Q 1 on the capacitor is (1-1/e) of its asymptotic charge Q f =C . –What is the relation between Q 1 and Q 2, the charge on the capacitor at time t=t 2 =2 ? q

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RC Circuits (Time-varying currents) Discharge capacitor: C initially charged with Q=C Connect switch to b at t=0. Calculate current and charge as function of time. Convert to differential equation for q: C a b R I I Loop theorem

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RC Circuits (Time-varying currents) S olution: Check that it is a solution: ! Note that this “guess” incorporates the boundary conditions: C a b R I I Discharge capacitor:

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RC Circuits (Time-varying currents) Current is found from differentiation: Conclusion: Capacitor discharges exponentially with time constant = RC Current decays from initial max value (= - /R) with same time constant Discharge capacitor: C a b + -- R + I I

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Discharging Capacitor Charge on C Max = C 37% Max at t=RC t q 0 CC RC 2RC 0 - R I t Current Max = - /R 37% Max at t=RC

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Lecture 12, CQ 3 At t=0 the switch is connected to position a in the circuit shown: The capacitor is initially uncharged. –At t = t 0, the switch is thrown from position a to position b. –Which of the following graphs best represents the time dependence of the charge on C? (a) (b) (c)

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At t=0 the switch is connected to position a in the circuit shown: The capacitor is initially uncharged. –At t = t 0, the switch is thrown from position a to position b. –Which of the following graphs best represents the time dependence of the charge on C? (a) (b) (c) For 0 < t < t 0, the capacitor is charging with time constant = RC For t > t 0, the capacitor is discharging with time constant = 2RC (a) has equal charging and discharging time constants (b) has a larger discharging than a charging (c) has a smaller discharging than a charging

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Charging Discharging RC t 2RC 0 - R I t q 0 CC RC t q 2RC 0 CC I 0 t R

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