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Fisika_soal dan pembahasan elektrik

02 Apr

QUESTIONS AND DISCUSSION

ABOUT THE DIRECTION OF ELECTRIC CURRENT CIRCUIT

1. A charge of 225 C flows in 30 minutes. What is the magnitude of the flowing current?

a. a. 0.25 A

b. b. 0.125 A

c. c. 0.5 A

d. d. 0.625 A

e. e. 1 A

Answer       :     t = 30 minutes = 1800 s

ΔQ = 225 C

I = = = 0.125 C/s = 0.125 A                                                               = b.

2.   A 2 m long wire has an area of 5 x10-4 m2. If its resistivity is 1.5×10-4 Ω, what is the resistance of the wire?

a. 0.8 Ω

b. 0.9 Ω

c. 0.6 Ω

d. 0.5 Ω

e. 0.7 Ω

Answer       :     l = 2 m

A      = 5 x10-4 m2

ρ       = 1.5×10-4

R      = ρ = 1.5×10-4 Ω x = 0.6 Ω                                                           = c

3.   Four resistors of 10 Ω, 25 Ω, 40 Ω, 35 Ω respectively. Are serially circuited. Determine the resistance of a replacement resistor (RP) to replace them.

a.  110 Ω

b. 120 Ω

c. 130 Ω

d. 50 Ω

e. 25 Ω

Answer       :     RP = R1+R2+R3+R4

= 10 Ω + 25 Ω + 40 Ω + 35 Ω

= 110 Ω                                                                                                               = a

4.   Three resistors of 10 Ω, 20 Ω, and 10 Ω respectively. Are pararelly circuited. Determine the resistance of a replacement resistor (RP) to replace them.

a.  35 Ω

b. 4 Ω

c. 10 Ω

d. 50 Ω

e. 25 Ω

Answer       : = + +

= + +

=

=

Rp = = 4 Ω                                                                                                        = b

5.   three resistors of 4 Ω, 6 Ω, and 12 Ω respectively. Are pararelly circuited. Determine the resistance of a replacement resistor (RP) to replace them.

a.  1 Ω

b. 2 Ω

c. 3 Ω

d. 4 Ω

e. 5 Ω

Answer       : = + +

= + +

=

=

Rp = = 2 Ω                                                                                                        = b

6. Ammeters are used to measure the strong currents that passes through a resistor was shown
value of 1.5 A. What is the charge that flows through the resistor in half a minute?
a. 40 C

b. 35 C

c. 25 C

d. 45 C

e. 50 C

Answer       :     I        = 1.5 A, t = 0.5 min = 30 s
Number of electric charge to meet:
Q         = I. t = 1.5. 30 = 45 C                                                                                        = d

7. Atthe ends ofa resistorgivenpotential difference 1.5volts. Whenstrongcurrentsweremeasuredat0.2 A.If apotential difference ofthe ends ofthe resistorchanged to4.5volts, what is the resistancecurrents measurable?
a. 8 Ω

b. 6.5 Ω

c. 7.5 Ω

d. 9 Ω

e. 5 Ω

Answer       :     V1 =1.5volts
I1 =0.2A

V2 =4.5volts
From thefirstconditioncan be obtained fromthe valuebarrier Rfor:
V1 =I1. R
1.5      =0.2.R
R                  =7.5Ω                                                                               = c

8. A Conductor is made of copper has 2 m length and 1.5 mm2 cross-sectional area. Obstacle Conductor is of 200 Ω. Conductor resistance, what is it?

a. 1,5.10-8Ωm

b. 1,5.10-9Ωm

c. 2.10-8Ωm

d. 2.10-9Ωm

e. 3.10-8Ωm

Answer :     l = 2 m, A = 1,5.10-6 m2, R = 200 Ω

R1          = ρ

200        = ρ. ρ                  = 1,5.10-8Ωm                                                             =a

9. Three resistance R1 = 20 Ω, R2 = 30 Ω and R3 = 50 Ω series strung together and connected to the potential difference 4.5 volts as in Figure…barriers tosetreplacement!

a. 120 Ωm

b. 110 Ωm

c. 100 Ωm

d. 90 Ωm

e. 80 Ωm

Answer                         :   Barriers replacement series satisfy:
Rs = R1 + R2 + R3 = 20 + 30 + 50 = 100 Ω                                               =c

10. Known three resistors are arranged in series with constraints for R1 = 100 Ω, R2 = 200 Ω
and R3 = 300 Ω. The ends of the circuit was connected the source voltage 120 volts. determine the  electric currents on the circuit!

a. 0.5 A           b. 0.25A                                    c. 0.75A                   d. 0.2 A              e. 1A

Answer    : R1 = 100 Ω, R2 = 200 Ω ,  R3 = 300 Ω

V      = 120 V

Rp =  R1 + R2 + R3 = 100 + 200 + 300 = 600

V      = I.R

I        = = = = 0.2 A                                                             =e

Figure from no 11-13

11.    Determine thereplacementofa seriesresistorom the figure!

a. 10 Ω

b. 12 Ω

c. 13.33 Ω

d. 16.66 Ω

e. 20 Ω

Answer   : = + +

= + +

=

=

Rp = = 13.33 Ω                                                                                               =c

12. Strong current through the resistance R2 is?

a. 1 Ω              b. 2 Ω                              c. 3 Ω                       d. 4 Ω                           e. 5 Ω

Answer    :     I1 = 2A

In a parallel series of different potential barriers same meaning applies the following relationship.
V2 = V1
I2. R2 = I1. R1
I2. 60 = 2.120
I2         = 4A                                                                                                                 =d

13.    Strong current through the resistance R3 is?

a. 12 Ω

b. 11 Ω

c. 13 Ω

d. 14 Ω

e. 15 Ω

Answer    :     In the same way can be strong currents I3.

V3 = V1 I3. R3 = I1. R1 I3. 20 = 2. 120           I3 = 12 A

=a

Figure for question  14-16 .

14 . Some obstacles arranged as in Figure above. If the ends are connected in different AB potential of 30 volts then specify substitute barriers AD!

a. 11 Ω

b. 12 Ω

c. 13 Ω

d. 14 Ω

e. 15 Ω

Answer       :   First    : to be determined so that the circuit RS1 be like Figure (a)

RS1 = 4 Ω + 6 Ω + 2 Ω = 12 Ω
Second: RS1 and 6 Ω arranged parallel means obtained

RBC = Rp and the composition of the Picture  (b).
= +
=
+ = =
RBC = Rp = 4 Ω
Third: Rp, 6 Ω and 5 Ω composed series means Rtotmeet:
Rtot = Rp + 6 + 5
= 4 + 6 + 5 = 15 Ω   =e

15.    Determine the potential difference VBC !

a. 7 V

b. 8 V

c. 9 V

d. 10 V

e. 11 V

Answer    :     Rtot value can be determined from strong currents through the circuit, which

satisfy:

I  = = = 2 A

So the potential difference can be obtained VBC

registration: VBC = I. RBC

= 2 . 4 = 8 volt                                                                               =b

16.    Strongspecifythe currentthroughresistance2Ω is ?

a. 0.75 A

b. 0.25 A

c. 0.5 A

d. 1 A

e. 2 A

Answer          : Strongcurrentspassing through thesameobstacles2Ω bywhichmeans it   .                                canpass throughbarriersRS1determined asfollows.

I= = =0.75A                                                                                =a

17.    A20Ωresistanceis connectedto thebattery Figure (b) 6voltvoltage. Specify the Absorbedpowerconstraints!

a. 1.75 Watts

b. 1.25 Watts

c. 1.5 Watts

d. 1.8Watts

e. 2 Watts

Answer                      :           R = 20 Ω        V = 6 volts      t = 0.5 min = 30 s

Absorbed power to meet:      P = > = 1.8 watts                     =d

18.    Resistance wire grid will increased in summer reasons:

(1) The wire becomes longer

(2) An electric current decreased during the day
(3) Inhibition type of wire increases

(4) Area cross section wire swell

The truth is ….

a. (1), (2) and (3)

b. (1) and (3)

c. (2) and (4)

d. (4)

e. all
Answer       :       B. (1) and (3)

19. Knownwithin1minute, ataflowConductor chargeof150coulombs. Howstrongcurrents flowingin Conductorthese?

a. 2 A

b. 3 A

c. 5 A

d. 2.5 A

e. 4 A
Given          :t=1min=60s
q=150C
Asked          : I=…?
Answer       : I=
= =2.5A
Thus, thestrongcurrentsflowingon theConductoris2.5A.                                                           =d

20.    To movecargo4coulombfrompoint A toBis required effortby 10joules. Determinethe  potential differencebetweenpointsAandB!

a. 2 V

b. 3 V

c. 2.5 V

d. 2.75 V

e. 4 V

Given          :q=4C W=10J
Asked          : V=…?
Answer       : V=
= = 2.5 V                                                                                                         =c

21. Knownstrongcurrentsof 0.5amperesflowsinaConductor whichhas a6voltpotential difference. Determinebarriers Conductorpoweris!

a. 12 b. 13 c. 12.5 d. 12.75 e. 14
Given          :V=6V               I=0,5 A
Asked          : R=…?
Answer       : V=I×R
à R= àR= à R= 12 Ω                                                                =a

22. Wirewhose length is200metersandcross section0.5mm2has56Ωelectrical resistance. DeterminebarriersThetype ofwire!

a. 1.4×10-8 Ω m.

b. 1.4×10-9 Ω m.

c. 1.4×10-6 Ω m.

d. 1.4×10-7 Ω m.

e. 1.4×10-5 Ω m.
Given       :a.l=200mb.A=0.5mm2=5×10-7m2c.R=56Ω
Asked       : ρ =…?
Answer    :ρ=
= = 1.4×10-7Ωm

Thus, the barrier type ofwireis1.4×10-7Ωm.                                                                 =d

Notethe pictureonthe side!

It is for question 23-25

23.    Ifthe huge flow ofincoming200 mA, thencalculatethe strength ofcurrentsI1.!

a. 120 mA

b. 130 mA

c. 140 mA

d. 150 mA

e. 160 mA

Answer    : Ientered =I1+I2

200 =I1+80

I1 =200-80

=120mA                                                                                                             =a

24.    Ifthe huge flow ofincoming200 mA, thencalculatethe strength of currentsI3.!

a. 50 mA

b. 60 mA

c. 70 mA

d. 80 mA

e. 90 mA

Answer    : I1 =I3+I4

120 =I5+40

I3 =120-400

=80mA                                                                                                                =d

25.    Ifthe huge flow ofincoming200 mA, thencalculatethe strength of currentsI5.!

a. 150 mA

b. 260 mA

c. 270 mA

d. 380 mA

e. 200 mA

Answer    : I5 =I2+I3 + I4

I5 =80+80 + 40

=200mA                                                                                                             =e

26.    Aconductivewireis connectedwithresistance11.5ohm  with6Vvoltage sourceiswithin0.5ohmsresistance. Calculate thecurrentsinthe circuit!

a. 1 A

b. 0.5 A

c. 2 A

d. 3 A

e. 1.5 A

Given       :a.R: 11.5      b.E: 6V       c.r=0.5

Asked :  I=…?
Answer    :Strongcurrentsin thecircuit
I =
= = 0.5 A                                                                                                 =b

Notethe pictureonthe side!

It is for question 27-29

27. Calculate thecurrentsinI1in thepicture above!

a. -1 A

b. 0.5 A

c. -2 A

d. 3 A

e. –1.5 A

Given       :  a.E1=8V       b.R1=4       c.E2=18V         d.R2=2             e.R3=6

Ask           : I1 = …….?

Answer          :

First Law of Kirchhoff
I3 = I1 + I2 ………… (1)

 

– Loop 1

ΣE     = Σ (I x R)
8        = I1 x 4 + I3 x 6
8        = I1 x 4 + (I1 + I2) 6
8 + 6  = 10 I1 + 6 I2 ……….. (2)
– Loop II
E2 = I2 x R2 + I3 x R3
18 = I2 x 2 + (I2 + I1) 6
18 = 2I2 + 6I1 + 6I2
9   = 3 I1 + 4I2 ……….. (3)
You elimination equation (2) and (3)

16  =  20 I1 + 12 I2

27  =  9 I1 + 12 I2

-11 = 11 I1

I1 = -1 A                                                                                                 =a

28. Calculate thecurrentsinI2in thepicture above!

a. 1 A

b. 0.5 A

c. 2 A

d. 3 A

e. 1.5 A

Answer    : The value I 1, you enter into the equation (2)
8 = 10 I1 + 6 I2
8 = 10 (-1) + 6 I2
I2 = 3 A                                                                                                                             =d

29. Calculate thecurrentsinI3in thepicture above!

a. 1 A

b. 0.5 A

c. 2 A

d. 3 A

e. 1.5 A

Answer    :   I2value, youenteredtheequation(1)

I3 =I1+I2

=-1+3

=2                                                                                 =c

30. There are three obstacles, each worth 6 Ω, 4 Ω, and 3 Ω in series. Determine barriers to replacement!

a. 13 A

b. 10.5 A

c. 12 A

d. 14 A

e. 15 A
Given    : a. R1 = 6 Ω          b. R2 = 4 Ω              c. R3 = 3 Ω
Asked    : Rs= … ?

 

Answer : Rs = R1 + R2 + R3

= 6 + 4 + 3

= 13 Ω
Thus, the barrier is 13 Ω successor.                                                                                              =a

 
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Posted by on April 2, 2011 in My School

 

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