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A copper wire of radius 0.1 mm and resistance 1 kW is connected across a power supply of 20 V. How many electrons are transferred per second between the supply and the wire at one end ?
- a)1.25 x 10-17
- b)1.25
- c)6.23 x 10-23
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A copper wire of radius 0.1 mm and resistance 1 kW is connected across...
To find the number of electrons transferred per second between the power supply and the wire, we can use the formula:
I = Q/t = V/R
where I is the current in the wire, Q is the charge transferred, t is the time over which the charge is transferred, V is the potential difference across the wire, and R is the resistance of the wire.
In this case, V is 20 V, R is 1 kW, and t is 1 second. We can rearrange the formula to solve for Q:
Q = I * t = V * t / R = 20 V * 1 s / 1 kW = 20 C
The charge transferred in one second is 20 C. Since the charge of an electron is 1.60 x 10^-19 C, the number of electrons transferred is:
Q / q = 20 C / (1.60 x 10^-19 C) = 1.25 x 10^17 electrons
Therefore, the answer is (a) 1.25 x 10^-17 electrons.
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Community Answer
A copper wire of radius 0.1 mm and resistance 1 kW is connected across...
Given:
Radius of copper wire, r = 0.1 mm = 0.1 × 10^(-3) m
Resistance of copper wire, R = 1 kW = 1 × 10^3 Ω
Voltage across the wire, V = 20 V
To find:
Number of electrons transferred per second between the supply and the wire at one end.
Formula Used:
Resistance of a wire, R = (ρ × L) / A
where ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
The resistivity of copper, ρ = 1.7 × 10^(-8) Ω·m (given)
The cross-sectional area of the wire, A = πr^2
Calculations:
Since we have the resistance and the radius of the wire, we can calculate the length of the wire using the formula for resistance.
R = (ρ × L) / A
1 × 10^3 = (1.7 × 10^(-8) × L) / (π × (0.1 × 10^(-3))^2)
Simplifying the equation:
1 × 10^3 = (1.7 × 10^(-8) × L) / (π × 0.01 × 10^(-6))
1 × 10^3 = (1.7 × 10^(-8) × L) / (3.14 × 10^(-8))
1 × 10^3 = 0.54 × L
L = (1 × 10^3) / 0.54
L = 1851.85 m
Now, we can calculate the cross-sectional area of the wire:
A = πr^2
A = π × (0.1 × 10^(-3))^2
A = π × (0.01 × 10^(-6))
A = 3.14 × 10^(-8) m^2
Now, we can calculate the current flowing through the wire using Ohm's Law:
V = IR
20 = I × 1 × 10^3
I = 20 × 10^(-3) A
I = 2 × 10^(-2) A
To find the number of electrons transferred per second, we need to calculate the charge flowing through the wire per second. The charge flowing per second can be calculated using the formula:
Q = I × t
where Q is the charge, I is the current, and t is the time.
Let's assume that t = 1 second.
Q = (2 × 10^(-2)) × 1
Q = 2 × 10^(-2) C
Now, we know that the elementary charge is 1.6 × 10^(-19) C. We can divide the total charge by the elementary charge to find the number of electrons:
Number of electrons = Q / (1.6 × 10^(-19))
Number of electrons = (2 × 10^(-2)) / (1.6 × 10^(-19))
Number of electrons = 1.25 × 10^17
Therefore, the number of electrons transferred per second between the supply and the wire at one end is 1
Radius of copper wire, r = 0.1 mm = 0.1 × 10^(-3) m
Resistance of copper wire, R = 1 kW = 1 × 10^3 Ω
Voltage across the wire, V = 20 V
To find:
Number of electrons transferred per second between the supply and the wire at one end.
Formula Used:
Resistance of a wire, R = (ρ × L) / A
where ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
The resistivity of copper, ρ = 1.7 × 10^(-8) Ω·m (given)
The cross-sectional area of the wire, A = πr^2
Calculations:
Since we have the resistance and the radius of the wire, we can calculate the length of the wire using the formula for resistance.
R = (ρ × L) / A
1 × 10^3 = (1.7 × 10^(-8) × L) / (π × (0.1 × 10^(-3))^2)
Simplifying the equation:
1 × 10^3 = (1.7 × 10^(-8) × L) / (π × 0.01 × 10^(-6))
1 × 10^3 = (1.7 × 10^(-8) × L) / (3.14 × 10^(-8))
1 × 10^3 = 0.54 × L
L = (1 × 10^3) / 0.54
L = 1851.85 m
Now, we can calculate the cross-sectional area of the wire:
A = πr^2
A = π × (0.1 × 10^(-3))^2
A = π × (0.01 × 10^(-6))
A = 3.14 × 10^(-8) m^2
Now, we can calculate the current flowing through the wire using Ohm's Law:
V = IR
20 = I × 1 × 10^3
I = 20 × 10^(-3) A
I = 2 × 10^(-2) A
To find the number of electrons transferred per second, we need to calculate the charge flowing through the wire per second. The charge flowing per second can be calculated using the formula:
Q = I × t
where Q is the charge, I is the current, and t is the time.
Let's assume that t = 1 second.
Q = (2 × 10^(-2)) × 1
Q = 2 × 10^(-2) C
Now, we know that the elementary charge is 1.6 × 10^(-19) C. We can divide the total charge by the elementary charge to find the number of electrons:
Number of electrons = Q / (1.6 × 10^(-19))
Number of electrons = (2 × 10^(-2)) / (1.6 × 10^(-19))
Number of electrons = 1.25 × 10^17
Therefore, the number of electrons transferred per second between the supply and the wire at one end is 1
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A copper wire of radius 0.1 mm and resistance 1 kW is connected across a power supply of 20 V. How many electrons are transferred per second between the supply and the wire at one end ?a)1.25 x 10-17b)1.25c)6.23 x 10-23d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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