Review Lecture 2. Copyright © 2009 Pearson Education, Inc. Admin: Mid-term is in class on Wednesday. No assignment this week. Labs and discussion sessions.

Review Lecture 2

Copyright © 2009 Pearson Education, Inc.

Admin:• Mid-term is in class on Wednesday.• No assignment this week. Labs and

discussion sessions run as usual

• Extra Office hours• 2-4pm Monday: Dr Holder Sharp Lab 222• 2-4pm Tuesday: Dr Holder Sharp Lab 222• Email is also OK (but visits work better)


What does the test cover?

• Electric fields and forces.• Potential of an infinite plane• Field between two plates• Capacitors

• Capacitance• Charge and energy on a capacitor• Combination rules• Capacitors in steady state

• Inductors• Inductance• Current and energy in an inductor• Combination rules

Ch. 21-5, -6, -7, -8

Ch. 24-1, -2, -3, -4

Ch. 30-2, -3


What does the test cover?• AC circuits

• AC waves• RMS quantities• Complex representation (including phasors)

• Components in AC circuits• Resistors• Capacitors (Reactance, impedance, i-v relation)• Inductors (Reactance, impedance, i-v relation)

• AC circuit analysis• Transient circuits

• Capacitor charging and discharging (RC circuits)• Inductor charging and discharging (LR circuits)• LC circuits – Oscillators• LRC circuits – Damped Oscillators

• Filters

Ch. 26-5

Ch. 30-4Ch. 30-5

Ch. 30-6


What formulae will I have?


What will be the format of the exam?

Similar to the last mid-term – some short answer, plus a few longer questions

Rules•Closed book exam – no notes or textbooks.•Please seat yourselves with at least one space between you.•A scientific calculator is required•Graphing calculators are OK – but I can’t give partial credit for the algebra if you don’t show it.•No phones/ phone calculators

•Students with accommodations can use the test center – but fill out the DSS form.


How should I study?

15% of your grade

The best resource is the lecture notes – in particular, make sure you can solve all of the worked examples, without the solutions in front of you.

The textbook is also useful.

Review the discussion problems, and homework problems.

Come to office hours (soon!) if anything is not clear – or drop me an email, but office hours are usually much more effective


24-2 Determination of CapacitanceExample 24-1: Capacitor calculations.

(a)Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm and are separated by a 1.0-mm air gap.

(b) What is the charge on each plate if a 12-V battery is connected across the two plates?


24-4 Electric Energy Storage

Example 24-8: Energy stored in a capacitor.

A camera flash unit stores energy in a 150-μF capacitor at 200 V. (a) How much electric energy can be stored? (b) What is the power output if nearly all this energy is released in 1.0 ms?


24-4 Electric Energy Storage

Conceptual Example 24-9: Capacitor plate separation increased.

A parallel-plate capacitor carries charge Q and is then disconnected from a battery. The two plates are initially separated by a distance d. Suppose the plates are pulled apart until the separation is 2d. How has the energy stored in this capacitor changed?

Remember that the charge is conserved

parallelseries

Capacitors

Inductorsparallelseries

All have the same charge All have the same voltage

C1 = 2.0 μF, C2 = 1.5 μF, and C3 = 3.0 μF, what arrangement will give the minimum voltage drop across the 2.0-μF capacitor C1?

Minimum voltage drop will be across maximum equivalent capacitance

C1 = 2.0 μF, C2 = 1.5 μF, and C3 = 3.0 μF, what arrangement will give the minimum voltage drop across the 2.0-μF capacitor C1?

If the battery provides 12V, what is the voltage drop across C1?

C13=5.0 μFCeq= C13C2/(C13 + C2)=1.15μF

Veq=12V

Q=VC=13.8μC

Q=13.8μCV13=Q/C13 =2.8VV13=V1

C13=5.0 μF

Capacitative Impedance

Inductive Impedance

Resistive Impedance

Impedance is the complex form of a component’s resistance

Current leads voltage by 90° in a capacitorCurrent has to flow first, to push charge onto the plates, which creates the voltage

Voltage leads current by 90° in an inductor

Voltage and current are in phase in a resistor


What is the rms current in the circuit shown if C = 1.0 μF and Vpeak = 170 V?

Calculate (a) for f = 60 Hz and then (b) for f = 6.0 x 105 Hz.

Note: we can simply use reactance (XC) if dealing with phase independent quantities like rms. But the answer is not valid at any given instant in time: for that we need the impedance (ZC)

Vrms = Vpeak/√2

ω = 2πf

Current Divider!

Sketch the input current and the current through the 1H inductance

ω = 2πf f=1/T

IL2= IS (REQ/R)


The capacitor is: Charging with the switch in position 1.Discharging with the switch in position 2.a) Will it take longer to charge or discharge?b) What is the current through the resistor at the start of the charging cycle?c) What is the maximum Charge on the capacitor?d) How long will it take to reach half of this maximum charge?

Kirchhoff’s loop rule.τ =RC

charging

Which of the following are high-pass or low-pass filters?

Answers: (b) and (c) are high-pass; (a) and (d) are low-pass

Review Lecture 2. Copyright © 2009 Pearson Education, Inc. Admin: Mid-term is in class on Wednesday. No assignment this week. Labs and discussion sessions.

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