Lecture 10

Lecture 10 - Hour Exam 1 Grade Histogram 30 25 20 15 10 5 0...

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0 5 10 15 20 25 30 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100 Class average 75 Hour Exam 1 Grade Histogram

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+ - i Consider the circuit shown in the figure. We assume that the capacitor is initially uncharged and that at 0 we throw the switch S from the middle position to posi t RC circuits : Charging of a capacitor tion a. The battery will charge the capacitor through the resistor . C R Our objective is to examine the charging process as function of time. We will write KLR starting at point b and going in the the counterclockwise direction. 0 The current q dq dq q iR i R C dt dt C If we rearrange the terms we have: This is an inhomogeneous, first order, linear differential equation with initial condition: (0) 0. This condition expresses the fact that a t dq q R dt C q t 0 the capacitor is uncharged. (27 15)
/ Differential equation: Intitial condition: (0) 0 Solution: 1 Here: The constant is known as the "time constant" of the circuit. If we plot versus we see t t dq q R dt C q q C e RC qt hat does not reach its terminal value but instead increases from its initial value and it reaches the terminal value at . Do we have to wait for an internity to charge the capacitor? In pract q C t / ice no. ( ) 0.632 ( 3 ) 0.950 ( 5 ) 0.993 If we wait only a few time constants the charge, for all practical purposes has reached its terminal value . The current If t q t C q t C q t C C dq ie dt R we plot i versus t we get a decaying exponential (see fig.b) (27 16) RC

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+ - i q t q o O Consider the circuit shown in the figure. We assume that the capacitor at 0 has charge and that at 0 we throw the switch S from the middle position to pos o tq t RC circuits : Discharging of a capacitor ition b. The capacitor is disconnected from the battery and looses its charge through resistor . We will write KLR starting at point b and going in the counterclockwise direction: 0 Taking int R q iR C o account that we get: 0 dq dq q iR dt dt C -/ This is an homogeneous, first order, linear differential equation with initial condition: (0) The solution is: Where . If we plot q versus t we get a decaying exponetial. t oo q q q q e RC The charge becomes zero at . In practical terms we only have to wait a few time constants. ( ) 0.368 , (3 ) 0.049 , (5 ) 0.007 o o o t q q q q q q (27 17) RC
The energy stored in a capacitor is equal to q 2 / 2 C . When a capacitor is discharged, what fraction of the initial energy remains after an elapsed time of one time constant? 1 2 3 4 5 0% 0% 0% 0% 0% 1. 1/e 2. 1/e 2 3. 1-1/e 4. (1-1/e) 2 5. Depends on how much energy was initially stored

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Lecture 10 Magnetic Field
New concepts Magnetic field vector Magnetic force on a moving charge Magnetic field lines Motion of a moving charge particle in a uniform magnetic field Magnetic force on a current carrying wire Magnetic torque on a wire loop Magnetic dipole, magnetic dipole moment Hall effect Cyclotron particle accelerator

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One can generate a magnetic field using one of the following methods: Pass a current through a wire and thus form what is knows as an "electromagnet".
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Lecture 10 - Hour Exam 1 Grade Histogram 30 25 20 15 10 5 0...

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