241-24_Lec-9 - Physics 241 Lecture 9 Y E Kim Chapter 24...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 241 Lecture 9 Y. E. Kim September 21, 2010 Chapter 24 September 21, 2010 University Physics, Chapter 23 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
September 21, 2010 University Physics, Chapter24 2 Review - So Far … The capacitance of a spherical capacitor is r 1 is the radius of the inner sphere r 2 is the radius of the outer sphere The capacitance of an isolated spherical conductor is R is the radius of the sphere 12 0 21 4 rr C  0 4 CR
Background image of page 2
September 21, 2010 University Physics, Chapter24 3 Review (2) The equivalent capacitance for n capacitors in parallel is The equivalent capacitance for n capacitors in series is 1 n eq i i CC 1 11 n i eq i = =
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
September 21, 2010 University Physics, Chapter24 4 Example: System of Capacitors (1) Question: If each capacitor has a capacitance of 5 n F , what is the capacitance of this system of capacitors? Answer: Find the equivalent capacitance Analyze each piece of the circuit individually, replacing pairs in series or in parallel by one capacitor with equivalent capacitance
Background image of page 4
September 21, 2010 University Physics, Chapter24 5 Example: System of Capacitors (2) We can see that C 1 and C 2 are in parallel, and that C 3 is also in parallel with C 1 and C 2 We find C 123 = C 1 + C 2 + C 3 = 15 n F … and make a new drawing
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
September 21, 2010 University Physics, Chapter24 6 Example: System of Capacitors (3) We can see that C 4 and C 123 are in series We find for the equivalent capacitance: … and make a new drawing 1 C 1234 1 C 123 1 C 4 C 1234 C 123 C 4 C 123 C 4 = 3.75 nF
Background image of page 6
September 21, 2010 University Physics, Chapter24 7 Example: System of Capacitors (4) We can see that C 5 and C 1234 are in parallel We find for the equivalent capacitance … and make a new drawing C 12345 C 1234 C 5 C 123 C 4 C 123 C 4 C 5 ( C 1 C 2 C 3 ) C 4 C 1 C 2 C 3 C 4 C 5 = 8.75 n F
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
September 21, 2010 University Physics, Chapter24 8 Example: System of Capacitors (5) So the equivalent capacitance of our system of capacitors Units and order-of-magnitude are ok More than one half of the total capacitance of this arrangement is provided by C 5 alone This result makes it clear that one has to be careful how one arranges capacitors in circuits. 12345 (5 5 5)5 5 nF 8.75 nF 5555 C     
Background image of page 8
Physics 241-24- Quiz 1 September 21,2010 Three capacitors are connected to a battery as shown in the figure. If C 1 = C 2 = C 3 = 10.0 μF and V = 10.0 V , what is the charge on capacitor C 3 ? a) 66.7 μC b) 100. μC c) 150. μC d) 300. μC e) 457. μC
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
September 21, 2010 University Physics, Chapter24 10 A battery must do work to charge a capacitor. We can think of this work as changing the electric potential energy of the capacitor. The differential work dW done by a battery with voltage V to put a differential charge dq on a capacitor with capacitance C is The total work required to bring the capacitor to its full charge q is This work is stored as electric potential energy Energy Stored in Capacitors q dW Vdq dq C  2 0 1 2 t q t t q q W dW dq CC  2 2 1 1 1 2 2 2 q U CV qV C
Background image of page 10
Image of page 11
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/28/2010 for the course PHY 222 taught by Professor Yu during the Spring '10 term at Beaufort County Community College.

Page1 / 31

241-24_Lec-9 - Physics 241 Lecture 9 Y E Kim Chapter 24...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online