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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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problem
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?
i w qsin w t f
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
li 2 q 2 2c q 2 2c
(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
problem5
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?
the inductor acts to oppose the source creating the change
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

using the loop rules
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
problem8
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?
part b first
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
part a the conceptual thing
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
problem11
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?
part a
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.
part b
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.
part c
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