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Chapter 24--Examples

Chapter 24--Examples. Problem. In the figure to the left, a potential difference of 20 V is applied across points a and b. What is charge on each capacitor if C 1 = 10 m F, C 2 =20 m F, and C 3 =30 m F. What is potential difference across points a and d? D and b?.

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Chapter 24--Examples

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  1. Chapter 24--Examples

  2. Problem In the figure to the left, a potential difference of 20 V is applied across points a and b. • What is charge on each capacitor if C1=10 mF, C2=20 mF, and C3=30 mF. • What is potential difference across points a and d? • D and b?

  3. Step 1: Find Equivalent Resistance • C1&C2: 10+20=30 • C1&C2+C3: (1/30)+(1/30)= 2/30 i.e. 30/2=15 mF

  4. Finding Potentials • Q=CV=15*20=300 mC • So charge on C3 is 300 mC • The voltage across C3 is 300 mC/30 mF= 10 V • So 10 V across points b & d and therefore, 20-10=10 V across a & d

  5. Parallel Network • If there is 10 V across this network then • Q1=10 mF*10V= 100 mC • If there are only 300 mC total and 100 mC is on this capacitor, then 200 mC must be charge of C2 • Check: Q2=20 mF*10V=200 mC

  6. Problem In the figure, each capacitor C1=6.9 mF and C2=4.6 mF. • Compute equivalent capacitance between a and b • Compute the charge on each capacitor if Vab=420 V • Compute Vcd when Vab=420 V

  7. Analysis • Look at the right most connection: it is parallel connection between the C2 capacitor and the 3 C1 capacitors • 3 C1’s: 6.9/3=2.3 mF • C2+3C1’s=2.3+4.6=6.9

  8. Repeat Again • Now the 6.9 mF capacitor is in series with the other C1 capacitors • So it is the same circuit as again, so the equivalent is 6.9 mF • Finally, the total equivalent is 2.3 mF

  9. So 420 V *2.3 mF= 966 mC • For each C1 capacitor on the leftmost network, the voltage across each is 140 V (966/6.9 or 420/3) • If there is 140 V across the C2, then 140*4.6=644 mC • There must be 966-644 = 322 mC in the other branch.

  10. In the middle network, • Each capacitor has 46.67 V (140/3 or 322 mC/6.9) • So voltage across c & d is 46.67 V • Then C2 capacitor has 214mC and the other C1 capacitors have 322-214 =107 mC

  11. Problem Two parallel plates have equal and opposite charges. When the space between them is evacuated, the electric field is 3.2 x 105 V/m. When the space is filled with a dielectric, the electric field is 2.5 x 105 V/m. • What is the charge density on each surface of the dielectric? • What is the dielectric constant?

  12. ETotal=E0-Ei ++++++++++++++++++++++ • - - - - - - - - - - - - - - - - - - • +++++++++++++++++++ E0 Ei - - - - - - - - - - - - - - - - - - - - - - - si=8.85e-12*(3.2-2.5)*105=4.32x106 C/m2

  13. Dielectric Constant • K=E0/E=3.2e5/2.5e5=1.28

  14. Problem A 3.4 mF capacitor is initially uncharged and then connected in series with a 7.25 kW resistor and an emf source of 180 V which has negligible resistance. • What is the RC time constant? • How much time does it take (after connection) for the capacitor to reach 50% of its maximum charge? • After a long time the EMF source is disconnected from the circuit, how long does it take the current to reach 1% of its maximum value?

  15. RC Value • RC=3mF*7.25kW • RC=3e-6*7.25e3 • RC=21.75 ms

  16. Time to 50% of max charge • Q(t)=C*V*(1-e-t/RC) • Q(t)/CV is the fraction of the maximum charge so let Q(t)/CV =50% • .5=1-e-t/RC • e-t/RC=.5=1/2 or 2-1 • -t/RC=-ln(2) • t=RC*ln(2)=21.75*.693 • t=15.07 ms

  17. Since R & C have not changed, RC=21.75 ms • I(t)=(V/R)*e-t/RC • I(t)/(V/R) is the fraction of maximum current • Let I(t)/(V/R) = 1% or 0.01 • 0.01=e-t/RC • ln(0.01)=-4.605=-t/RC • t=21.75*4.605=100 ms

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