chapter 30-suggested problems

8/13/2019 Chapter 30-Suggested Problems

1/5

30.11.IDENTIFY and SETUP: Apply / . L di dt Apply Lenz’s law to determine the direction of the induced emf in the

coil.

EXECUTE: (a) 3/ (0 260 H)(0 0180 A/s) 4 68 10 V L di dt

(b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a

battery it would have the terminal at a

EVALUATE: The induced emf is directed so as to oppose the decrease in the current.

30.26.IDENTIFY: With 1S closed and 2S open, ( )i t is given by Eq. (30.14). With 1S open and 2S closed, ( )i t is given

by Eq. (30.18).

SETUP: 212

.U Li After 1S has been closed a long time, i has reached its final value of / . I R

EXECUTE: (a) 21

2U LI and

2 2(0 260 J)2 13 A.

0 115 H

U I

L

(2 13 A)(120 ) 256 V. IR

(b) ( / ) R L t i Ie and 2 2 2( / ) 21 1 1 1 102 2 2 2 2 . R L t U Li LI e U LI 2( / ) 1

2,

R L t e so

41 12 20.115 H

ln ln 3.32 10 s.2 2(120 )

Lt

R

EVALUATE: 4/ 9.58 10 s. L R The time in part (b) is ln (2)/2 0347 .

30.27. IDENTIFY: Apply the concepts of current decay in an R- L circuit. Apply the loop rule to the circuit. ( )i t

is given by Eq. (30.18). The voltage across the resistor depends on i and the voltage across the inductor depends

on / .di dt

SETUP: The circuit with 1S closed and 2S open is sketched in Figure 30.27a.

0di

iR Ldt

Figure 30.27a

Constant current established means 0.di

dt

EXECUTE: 60.0 V

0.250 A240

i R

(a) SETUP: The circuit with 2S closed and 1S open is shown in Figure 30.27b.

( / )0

R L t i I e

At 00, 0 250 At i I

Figure 30.27b

The inductor prevents an instantaneous change in the current; the current in the inductor just after 2S is closedand 1S is opened equals the current in the inductor just before this is done.

(b) EXECUTE: 4( / ) (240 / 0 160 H)(4 00 10 s) 0 600

0 (0 250 A) (0 250 A) 0 137 A R L t i I e e e

(c) SETUP: See Figure 30.27c.


2/5


3/5

EVALUATE: At the end of 1.30 ms, nearly all the energy is now in the inductor, leaving very little in the

capacitor.

30.33.IDENTIFY: The energy moves back and forth between the inductor and capacitor.

(a) SETUP: The period is1 1 2

2 ./2

T LC f

EXECUTE: Solving for L gives2 5 2

2

2 2 9

(8 60 10 s)2 50 10 H 25 0 mH

4 4 (7 50 10 C)

T L

C

(b) SETUP: The charge on a capacitor is .Q CV

EXECUTE: 9 –8

(7.50 10 F)(12.0 V) 9.00 10 CQ CV


4/5

(c) SETUP: The stored energy is2/2 .U Q C

EXECUTE: 8 2

7

9

(9 00 10 C)5 40 10 J

2(7 50 10 F)U

(d) SETUP: The maximum current occurs when the capacitor is discharged, so the inductor has all the initial

energy. Total. L C U U U 21

Total20 . LI U

EXECUTE: Solve for the current:

73Total

2

2 2(5 40 10 J)6 58 10 A 6 58 mA

2 50 10 H

U I

L

EVALUATE: The energy oscillates back and forth forever. However, if there is any resistance in the circuit, no

matter how small, all this energy will eventually be dissipated as heat in the resistor.

30.66.IDENTIFY: At all times 1 2 25 0 V.v v The voltage across the resistor depends on the current through it and the

voltage across the inductor depends on the rate at which the current through it is changing.

SETUP: Immediately after closing the switch the current through the inductor is zero. After a long time the

current is no longer changing.

EXECUTE: (a) 0i so1

0v and2

250 V.v The ammeter reading is 0. A

(b) After a long time, 2 0v and 1 25 0 V.v 1v iR and1 25 0 V 1 67 A.

15 0

vi

R

The ammeter reading is

1 67A. A (c) None of the answers in (a) and (b) depend on L so none of them would change.

EVALUATE: The inductance L of the circuit affects the rate at which current reaches its final value. But after a

long time the inductor doesn’t affect the circuit and the final current does not depend on L.

30.68.IDENTIFY: Closing 2S and simultaneously opening 1S produces an L-C circuit with initial current through the

inductor of 3.50 A. When the current is a maximum the charge q on the capacitor is zero and when the charge q

is a maximum the current is zero. Conservation of energy says that the maximum energy21max2

Li stored in the

inductor equals the maximum energy2max1

2q

C stored in the capacitor.

SETUP: max 3 50 A,i the current in the inductor just after the switch is closed.

EXECUTE: (a) 2

2 max1 1max2 2

.q

LiC

3 6 4max max( ) (2 0 10 H)(5 0 10 F)(3 50 A) 3 50 10 C 0 350 mC.q LC i

(b) When q is maximum, 0.i

30.70.IDENTIFY: Apply the loop rule to the two loops. The current through the inductor doesn’t change abruptly.

SETUP: For the inductordi

Ldt

and is directed to oppose the change in current.

EXECUTE: (a) Switch is closed, then at some later time

50 0 A/s (0 300 H)(50 0 A/s) 15 0 V.cd di di

v Ldt dt

The top circuit loop: 60.0 1 1 1600V

V 1 50 A.40 0

i R i

The bottom loop: 2 2 2450V

60.0 V 15 0 V 0 1 80 A.25 0

i R i


5/5

(b) After a long time: 2600V

2 40 A,25 0

i

and immediately when the switch is

opened, the inductor maintains this current, so 1 2 2 40 A.i i

chapter 30-suggested problems

Documents