QUESTIONS AND DISCUSSION ABOUT THE DIRECTION OF ELECTRIC CURRENT CIRCUIT

Posted: March 18, 2011 in MASHUDI

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. 0.25 A

b.0.125 A

c. 0.5 A

d. 0.625 A

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

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