Chapter 30--Examples

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# Chapter 30--Examples - PowerPoint PPT Presentation

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

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=-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
(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
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
• 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
• 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
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!
• 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
• 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
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)
• 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)
• 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)
• 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