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The resistance of a wire of iron is 10 ohms and temp. coefficient of resistivity is 5x10-3∕℃. At 20oC it carries 30 milliamperes of current. Keeping constant potential difference between its ends, the temperature of the wire is raised to 120oC. The current in milliamperes that flows in the wire is
- a)20
- b)15
- c)10
- d)40
Correct answer is option 'A'. Can you explain this answer?
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The resistance of a wire of iron is 10 ohms and temp. coefficient of r...
Given:
Resistance of iron wire (R) = 10 ohms
Temperature coefficient of resistivity (α) = 5x10^-3
Current flowing through wire at 20°C (I1) = 30 milliamperes
Change in temperature (ΔT) = 120°C - 20°C = 100°C
To find:
Current flowing through wire at 120°C (I2)
Solution:
1. Using the formula for temperature dependence of resistance:
R2 = R1 (1 + α ΔT)
where R1 is the resistance at temperature T1, R2 is the resistance at temperature T2, α is the temperature coefficient of resistivity, and ΔT is the change in temperature.
2. Substituting the given values:
R2 = 10 ohms (1 + 5x10^-3 x 100)
R2 = 10 ohms (1.5)
R2 = 15 ohms
3. Using Ohm's law:
V = IR
where V is the potential difference between the ends of the wire, I is the current flowing through the wire, and R is the resistance of the wire.
4. Since the potential difference is kept constant, we can write:
V = I1 R1 = I2 R2
5. Substituting the given values:
I2 = I1 R1 / R2
I2 = 30 milliamperes x 10 ohms / 15 ohms
I2 = 20 milliamperes
6. Converting to the required units:
I2 = 20 milliamperes = 20 mA
Answer:
The current flowing through the wire at 120°C is option (a) 20 mA.
Resistance of iron wire (R) = 10 ohms
Temperature coefficient of resistivity (α) = 5x10^-3
Current flowing through wire at 20°C (I1) = 30 milliamperes
Change in temperature (ΔT) = 120°C - 20°C = 100°C
To find:
Current flowing through wire at 120°C (I2)
Solution:
1. Using the formula for temperature dependence of resistance:
R2 = R1 (1 + α ΔT)
where R1 is the resistance at temperature T1, R2 is the resistance at temperature T2, α is the temperature coefficient of resistivity, and ΔT is the change in temperature.
2. Substituting the given values:
R2 = 10 ohms (1 + 5x10^-3 x 100)
R2 = 10 ohms (1.5)
R2 = 15 ohms
3. Using Ohm's law:
V = IR
where V is the potential difference between the ends of the wire, I is the current flowing through the wire, and R is the resistance of the wire.
4. Since the potential difference is kept constant, we can write:
V = I1 R1 = I2 R2
5. Substituting the given values:
I2 = I1 R1 / R2
I2 = 30 milliamperes x 10 ohms / 15 ohms
I2 = 20 milliamperes
6. Converting to the required units:
I2 = 20 milliamperes = 20 mA
Answer:
The current flowing through the wire at 120°C is option (a) 20 mA.
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The resistance of a wire of iron is 10 ohms and temp. coefficient of resistivity is 5x10-3. At 20oC it carries 30 milliamperes of current. Keeping constant potential difference between its ends, the temperature of the wire is raised to 120oC. The current in milliamperes that flows in the wire isa)20b)15c)10d)40Correct answer is option 'A'. Can you explain this answer?
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