Type Barriers: Definition, Formulas, Values and Example Problems
Type Barriers: Definition, Formulas, Values and Example Problems – What is the type resistance and how is it calculated? On this occasion About the knowledge.co.id will discuss it and of course about other things that also cover it. Let's look at the discussion together in the article below to better understand it.
Type Barriers: Definition, Formulas, Values and Example Problems
Resistance is the resistance of a material to an electric current. The magnitude of the strong electric current flowing in a conductor is proportional to the voltage difference.
Resistance is the tendency of a material to resist the flow of electric current, denoted by ρ (rho). The determining factor for the size of the resistivity value of a conductor is the material of the conductor wire
We might think that the resistance possessed by a thick wire is smaller compared to thin wire, because thicker wire has a wider area to electron flow. We would also expect that the longer a conductor is, the greater the resistance, since there will be more of a barrier to the flow of electrons.
When an electric current or free electrons flow in a conductor that has a large diameter (section), the resistance value will be lower. This is because with a large diameter, the flowing electric current will be easier, but if the diameter of the conductor is small, the flowing electric current will be hampered.
Likewise, if the electric current flows over a longer distance, the value of the resistance will be even greater. but if the electric current flows at a shorter distance, the resistance value will also increase small.
Value and Type Resistance Formula
From the explanation above it can be concluded that the electrical resistance of a conductor will be proportional is directly proportional to the length of the conductor and inversely proportional to the cross-sectional area conductor.
Therefore systematically can be formulated with the following formula:
R = ρ x l/A
Information :
- R is the electrical resistance with units of Ω
- ρ is the resistivity with units of Ωm
- l is the length of the conductor in m
- A is the cross-sectional area in m2
The amount of ρ (read rho) depends on the specific resistance of the type of conductor used. The resistivity values of several types of conductors can be seen in the table below:
Examples of Type Obstacles
Problem 1
What is the value of the resistance of an iron wire conductor (type resistance 9.71 x 10 -8 Ωm) which has a wire length of 20 m and a diameter of 10 mm.
Resolution:
Is known :
ρ = resistivity 9.71 x 10 -8 Ωm
l = 20 m
D=10mm
Wanted: R = …. ?
Answer :
The first step is to first find the cross-sectional area of the iron wire. The cross-sectional area can be found by the formula:
A = ¼ x π x D2
A = = ¼ x 3.14 x 102
A = ¼ x 3.14 x 100
A = 78.5mm2 = 78.5 x 10-6 m2
So that the resistance of the iron wire conductor is:
R = ρ x l/A
R = 9.71 x 10 -8 Ωm x 20m / 78.5 x 10-6 m2
R = 2.47388 x 10-2 Ω
The wires in an incandescent lamp have a resistance of 20 Ω. If the lamp is connected to a voltage of 220 volts, what is the current flowing in the circuit?
Answer :
Noted that :
R = 20 Ω
V = 220 volts
R = V/i
20 = 220/i
i = 220/20 = 11 A
Problem 2
A copper wire is 1 m long and has a cross-sectional area of 5 mm2. If at 20 oC has a type resistance of 1.7. 10-6 Ωm and the temperature coefficient is 4. 10-3oC. How big is the resistance of the wire?
Answer :
Noted that :
l = 1 m
A = 5mm2
ρt = 1,7. 10-6 Ωm
α = 4. 10-3 o'clockC
t = 20 oC
Rt = ρt. la
= 1,7. 10-6. 1/5
= 0,34. 10-6 = 3,4. 10-5 Ω
Problem 3
A wire is 10 m long and has a cross-sectional area of 1 cm2 and type 10 resistance-5 Ωm. If the wires are made of the same material, have the same mass and have twice the cross-sectional area of the original wire, what is the resistance?
Answer :
Noted that :
l = 10 m
A1 = 1 cm2 = 0.0001 m2 = 10-4 m2
A2 = 2. 10-4 m2
ρ = 10-5 Ωm
R = ρ. la
=10-5. 10/2. 10-4
= 10-5. 5 .104 = 0,5 Ω
Problem 4
A piece of nichrome wire 3 meters long has a resistance of 20 ohms. The second nichrome wire is the same length, but the diameter is ½ the diameter of the first wire. What is the resistance of the second wire?
Completion:
knowni:
l1 = l2 = 3 m
d2 = ½ d1
R1 = 20 Ω
ρ1 = ρ2
asked:R2 = … ?
Answer:
Because the diameter d2 = ½ d1 then the radius of the wire is also the same, namely r2 = ½ r1. First find the cross-sectional area (A) of the second nikron wire using the formula for the area of a circle, namely:
L = πr2 so
L1 = πr2
L2 = π(½ r1)2 => L2 = ¼ πr12 => L2 = ¼L1
So, A2 = ¼A1
The resistance of the second type of conductor can be found by using the formula:
R = ρl/A
ρl = R.A
In this case the length and resistance of the wire types are the same, therefore:
(ρl)1 = (ρl)2
R1A1 = R2A2
20 ΩA1 = R2x ¼A1
R2 = 4 x 20 Ω
R2 = 80 Ω
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