No. 1
Within two minutes of electrical current of 2 A flows along the wires. Specify:
a) load switch
b) the number of electrons
Discussion
Strong relationships electric current, electric charge and time are:
I = Q / t
Q = I x t
Therefore :
a) Q = I x t
Q = 2 x 120
Q = 240 Coulomb
b) determine the number of electrons in the charge
n = Q / Qe
Where:
n = number of electrons
Qe = charge of the electron (1.6 x 10-19 Coulomb)
Q = charge will be calculated the number of electrons
so:
n = Q / Qe
n = 240 / (1.6 x 10-19)
n = 150 x 1019
n = 1.5 x 1021 electrons fruit
No. 5
A wire conductor has a length L and cross-sectional area A and has a resistance of 120 Ω. If the wire with the same material has a length of 2 L and 3 cross-sectional area A, define this second wire barrier!
Discussion
The formula for calculating the resistance of a wire conductor is:
R = ρL / A
Where
R = the resistance of wire (Ω)
L = length of wire (m)
A = cross-sectional area of the wire (m2)
ρ = resistivity wire
Wire made from the same have the same resistivity, so
R2 / R1 = (L2 / A2) / (L1 / A1)
R2 = (L2 / L1) x (A1 / A2) x R1
R2 = (2L / L) x (A / 3A) x 120
R2 = (2/1) x (1/3)
R2 = (2/3) x 120 = 80 Ω
Problem No. 2
A charge of 360 C flowing in two minutes in a circuit. Determine the current strength of the electrical circuit!
Discussion
I = Q / t
I = 360 coulombs / 120 second
I = 3 Ampere
Problem No. 3
Convert the following units:
a) 5 μ C = C ........
b) 100 mC = ......... C
c) 5000 nC = ......... C
Discussion
a) 5 μC = 5 x 10-6 = 0.000005 C
b) 100 mC = 100 x 10-3 = 0.1 C
c) 5000 nC = 5,000 x 10-9 = 5 x 10-6 = 0.000005 C
Problem No. 4
Convert the following units:
a) 5 kΩ = ....... Ω
b) 0.3 kΩ = ..... Ω
c) 5 MΩ Ω = .........
Discussion
a) 5 kΩ = 5 x 1000 = 5000 Ω
b) 0.3 kΩ = 0.3 x 1000 = 300 Ω
c) 5 MΩ = 5 x 1 million = 5000000 Ω
Problem No. 6
A conductive wire with a length of 10 meters has a resistance of 100 Ω If the wire is cut into two parts of equal length, determine the barriers owned by each piece of wire!
Discussion
Data of the matter:
L1 = L
L2 = 1/2 L
A1 = A2
R1 = 100 Ω
R2 / R1 = (L2 / L1) x R1
(Cross-sectional area and resistivity both wires are the same !!)
R2 = (0.5 L / L) x 100 Ω:
R2 = 50 Ω (Thanks to nasha for his input, ..)
Problem No. 7
Change the following units:
a) 300 mA = ......... A
b) 12 mA = .......... A
Discussion
a) 300 mA = 300: 1000 = 0.3 A
b) 12 mA = 12: 1000 = 0.012 A
Problem No. 8
Given a direct current electrical circuit consists of three lamps, two switches and a source of electric current. Where the lights are lit if:
a) switch 1 is closed, the switch 2 is open
b) switch 2 is closed, the switch 1 is open
c) switch 1 is closed, the switch 2 is closed
d) switch 1 is open, the switch 2 is open
Discussion
Electric current will flow if there is a load and the circuit is closed, so that:
a) switch 1 is closed, the switch 2 is open
2 and 3 lit the lamp, the lamp 1 dead
b) switch 2 is closed, the switch 1 is open
all the lights will die
c) switch 1 is closed, the switch 2 is closed
all the lights on
d) switch 1 is open, the switch 2 is open
all lights off
Problem No. 9
X and Y are two different electrical measuring devices. Which installation position voltmeter and ammeter if the measured voltage on the lamp 3 and strong currents at the lamp 3?
Discussion
Voltmeter to measure the voltage between two points, in this case the voltage on the lamp 3, voltmeter should be installed in parallel with the load to be measured, the correct position is X.
Ammeter to measure the electric current flowing through a load in this case is the lamp 3, ampermeter must be installed in series with a great tool to be measured strong currents, so, the cord cut before then sambungin to foot ammeters, the correct position is Y.
Problem No. 10
Consider the following picture
Determine the replacement barriers between points A and B if R1, R2 and R3 respectively magnitude is 5 Ω, 10 Ω and 15 Ω!
Discussion
Above circuit is composed of the series, so the barriers successor is:
Rt = R1 + R2 + R3
= 5 + 10 + 15
= 30 Ω
Problem No. 11
In the following series if R1 = 9 Ω and substitute barriers between point AB is a huge barrier 6 Ω specify R2!
Discussion
Above circuit is arranged paralal barriers. The total resistance for a parallel is
1 / Rt = 1 / R1 + 1 / R2
1/6 = 1/9 + 1 / R2
1 / R2 = 1/6 - 1/9
1 / R2 = 3/18 - 2/18
1 / R2 = 1/18
R2 = 18 Ohm
Problem No. 12
From the series following barriers R1 = 10 Ω
R2 = 9 Ω
R3 = 18 Ω
Determine the replacement barriers between points A and B!
Discussion
The series of mix, complete the parallel between R2 and R3 advance
1 / Rp = 1 / R2 + 1 / R3
1 / Rp = 1/9 + 1/18
1 / Rp = 2/18 + 1/18
1 / Rp = 3/18
Rp = 18/3 = 6 Ω
These results are then in series with R1
Rt = 10 + 6 = 16 Ω
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