Chapter 01-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

This is solution to problems related Electrical Circuit Analysis course. It was given by Prof. Gurnam Kanth at Punjab Engineering College. Its main points are: Coulombs, Current, Charge, Flow, Conductor, Charge, Time, Interval, Wire, Waveforms,

Typology: Exercises

2011/2012

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Chapter 1, Problem 1
How many coulombs are represented by these amounts of electrons:
(a) (b)
17
10482.6 ×18
1024.1 ×
(c) (d)
19
1046.2 ×20
10628.1 ×
Chapter 1, Solution 1
(a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C
(b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C
(c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C
(d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C
Chapter 1, Problem 2.
Determine the current flowing through an element if the charge flow is given by
(a)
() ( )
mC 83 += ttq
(b)
()
C 2)48 2t-t(tq +=
(c)
()
(
)
nC e5e3tq t2-t
=
(d)
()
pCt sin10 120tq
π
=
(e)
()
Ct 50cos20 4
μ
t
etq
=
Chapter 1, Solution 2
(a) i = dq/dt = 3 mA
(b) i = dq/dt = (16t + 4) A
(c) i = dq/dt = (-3e-t + 10e-2t) nA
(d) i=dq/dt =
1200 120
π
π
cos
t
pA
(e) i =dq/dt =
−+
et t
t4
80 50 1000 50
(cos sin )
A
μ
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Chapter 1, Problem 1

How many coulombs are represented by these amounts of electrons:

(a) (b)

17

  1. 482 × 10

18

  1. 24 × 10

(c) (d)

19

  1. 46 × 10

20

  1. 628 × 10

Chapter 1, Solution 1

(a) q = 6.482x

17 x [-1.602x

  • C] = -0.10384 C

(b) q = 1. 24x

18 x [-1.602x

  • C] = -0.19865 C

(c) q = 2.46x

19 x [-1.602x

  • C] = -3.941 C

(d) q = 1.628x

20 x [-1.602x

  • C] = -26.08 C

Chapter 1, Problem 2.

Determine the current flowing through an element if the charge flow is given by

(a) q ( ) t = ( 3 t + 8 ) mC

(b) ( ) 8 4 2)C

2 q t = ( t + t-

(c) q( ) t ( 3 e 5 e )nC

-t − 2 t = −

(d) q ( ) t = 10 sin 120 πtpC

(e) ( ) 20 cos 50 t C

4

t q t e

Chapter 1, Solution 2

(a) i = dq/dt = 3 mA

(b) i = dq/dt = (16t + 4) A

(c) i = dq/dt = (-3e

-t

  • 10e

-2t ) nA

(d) i=dq/dt = 1200 πcos 120 π t pA

(e) i =dq/dt = − + − e t t 4 t ( 80 cos 50 1000 sin 50 ) μ A

Find the charge q ( t ) flowing through a device if the current is:

(a) i ( ) t = 3 A, q ( ) 0 = 1 C

(b) i ( t )= ( 2 t + 5 )mA, q ( 0 )= 0

(c) i ( t )= 20 cos( 10 t + π / 6 ) μA, q (0)= 2 μC

(d) ( ) 10 sin 40 A, (0) 0

30 = =

i t e t q

t

Chapter 1, Solution 3

(a) = + =(3t +1) C ∫

q(t) i(t)dt q(0)

(b) (t 5t) mC

2 = + + = + ∫

q(t) (2t s)dt q(v)

(c) q(t) = 20 cos (^) (10t + π/ 6) + q(0) = (2sin(10 t + π/ 6) +1) μC ∫

(d)

e (0.16cos40t 0.12sin40t) C

  • 30t =− +

∫ ( 30 sin 40 t- 40 cost) 900 1600

10 e q(t) 10e sin 40 t q(0)

-30t

  • 30t

Chapter 1, Problem 4.

A current of 3.2 A flows through a conductor. Calculate how much charge passes

through any cross-section of the conductor in 20 seconds.

Chapter 1, Solution 4

q = it = 3.2 x 20 = 64 C

Chapter 1, Problem 5.

Determine the total charge transferred over the time interval of 0 ≤ t ≤ 10s when

1 ( ) 2

i t = t A.

Chapter 1, Solution 5

(^10 )

0

25 C

t q = idt = tdt = = ∫ ∫

The charge flowing in a wire is plotted in Fig. 1.24. Sketch the corresponding

current.

Figure 1.

Chapter 1, Solution 7

25A, 6 t 8

  • 25A, 2 t 6

25 A, 0 t 2

dt

dq i

which is sketched below:

The current flowing past a point in a device is shown in Fig. 1.25. Calculate the

total charge through the point.

Figure 1.

Chapter 1, Solution 8

10 1 15 μ C 2

q idt + × =

×

A rechargeable flashlight battery is capable of delivering 85 mA for about 12 h.

How much charge can it release at that rate? If its terminals voltage is 1.2 V, how

much energy can the battery deliver?

Chapter 1, Solution 11

q= it = 85 x

  • x 12 x 60 x 60 = 3,672 C

E = pt = ivt = qv = 3672 x1.2 = 4406.4 J

Chapter 1, Problem 12.

If the current flowing through an element is given by

( )

0,t 15s

  • 12 A, 10 t 15s

18 A, 6 t 10s

3 t A, 0 t 6s

it

Plot the charge stored in the element over 0 < t < 20s.

Chapter 1, Solution 12

For 0 < t < 6s, assuming q(0) = 0,

q t idt q tdt t

t t

( ) = + ( ) = + =. ∫ ∫

0 3 0 1 5

0

2

0

At t=6, q(6) = 1.5(6)

2 = 54

For 6 < t < 10s,

q t idt q dt t

t t

( ) = + ( )= + = − ∫ ∫

6 18 54 18 5

6 6

At t=10, q(10) = 180 – 54 = 126

For 10 At t=15, q(15) = -12x15 + 246 = 66

For 15