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Chapter 30--Examples

Chapter 30--Examples. Problem. In an L-C circuit, L =85 mH and C= 3.2 m F. During the oscillations the maximum current in the inductor is 0.85 mA What is the maximum charge on the capacitor?

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Chapter 30--Examples

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  1. Chapter 30--Examples

  2. Problem In an L-C circuit, L=85 mH and C=3.2 mF. During the oscillations the maximum current in the inductor is 0.85 mA • What is the maximum charge on the capacitor? • What is the magnitude of the charge on the capacitor at an instant when the current in the inductor has a magnitude of 0.5 mA?

  3. i=-wQsin(wt+f) • At maximum, sin()=1, so • i=-wQ • Where Q is the maximum value • w= • Q=0.85e-3/1917 • Q=0.44 mC

  4. (½ Li2)+(q2/2C)=(Q2/2C) • We know • L=85 mH • C=3.2 mF • Q=4.4e-7 C • We need to solve for q when i=0.5mA • q=3.58e-7 C

  5. Problem • In the circuit to the right, EMF=60 V, R1=40 W, R2=25 W, and L=0.30 H • Switch S is closed. At some time t afterward, the current in the inductor is increasing at a rate of di/dt=50 A/s. At this instant what is the current thru R1 and R2 ? • After the switch has been closed a long time, it is opened again. Just after it is opened, what is the current thru R1?

  6. The inductor acts to oppose the source creating the change • So we can replace the inductor with an EMF device, wired to oppose the 60 V device with an EMF= L*di/dt =0.3*50=15 V 60 V 40 W 15 V 25 W

  7. Using the loop rules • I1=60V/40 W = 1.5 A • I2=(60-15)/25 W =1.8 A • Part B) • After a long time, the current becomes steady, and di/dt=0 so I2=60/25=2.4 A • At the instant the switch is opened, the inductor maintains this current so I2=I1=2.4 A

  8. Problem • In the circuit, switch S is closed at t=0. • Find the reading of each meter just after S is closed. • What is the reading of each meter along time after S is closed?

  9. Part B) First! • After a long time, the inductors act like regular wire. • So we add the 5, 10, 15 W resistors in parallel to get 2.72 W • itotal =25 V/ (40+2.72 W )=0.585 A so A1=0.585 A • The voltage across the parallel circuit is 0.585*2.72=1.57 V • A2=1.57/5 = .314 A • A3=1.57/10=.157 A • A4=1.57/15=.104 A

  10. Part A– The conceptual thing • At the very instant that S is closed, inductors are maximally resisting. So they essentially PREVENT current flow through the 5 W and 10 W resistor! • A2=A3=0 • Thus A1 must equal A4 which is 25 V/ (40+15 W) =0.455 A

  11. Problem In the circuit, the switch S has been open for a long time and is suddenly closed. Neither the battery nor the inductors have any appreciable resistance. • What do the ammeter and voltmeter read just after S is closed? • What do the ammeter and the voltmeter after S has been closed a very long time? • What do the ammeter and the voltmeter read 0.115 ms after closing S?

  12. Part A) • At the very instant the switch is closed, the inductors act to stop current flow so A=0 • However, the potential difference is the full 20 V.

  13. Part B) • Again, after a long time, the inductors just act like wires so • i=20 V/(50+25 W) =0.267 A • V=0 since no potential difference.

  14. Part C) • To add the inductors, note that the 12 mH and 15 mH inductors are in series so L1215=27 mH • L1215 is in parallel with the 18 mH so • 1/LTot= (1/18)+(1/27) • LTot=10.8 mH • RTot=75 W • i=(EMF/RTot)*(1-e-(R/L)*t) • i=(20/75)*(1-e-(75/10.8e-3)*0.115e-3)=0.147 A • VR =iR=0.147*75=11 V • So potential difference across the inductor network must be 20-11= 9V

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