Question

A copper coil has a resistance of 20.0 ohm
at 0 degree celsius & resistance of 26.4 Ohm at
80 degree celsius . Find the temperature
Cofficent of resistance of copper.​

Answer 1

Answer:

The temperature coefficient is 4 × 10⁻³/°C

Explanation:

Given :

A copper coil has a resistance of 20 Ω at 0°C and resistance of 26.4 Ω at 80°C

To find :

the temperature coefficient of resistance of the wire

Solution :

Let α be the temperature coefficient of the wire.

We know,

$\underline{\boxed{\sf \alpha =\dfrac{R_2-R_1}{R_1(t_2-t_1)}}}$

where

R₁ denotes the resistance at temperature t₁

R₂ denotes the resistance at temperature t₂

Substitute,

R₁ = 20 Ω

R₂ = 26.4 Ω

t₁ = 0°C

t₂ = 80°C

$\tt \alpha = \dfrac{26.4-20}{20(80-0)} \\ \tt \alpha = \dfrac{6.4}{20(80)} \\ \tt \alpha = \dfrac{6.4}{1600} \\ \tt \alpha = 4 \times 10^{-3} / ^{\circ} C$

The temperature coefficient is 4 × 10⁻³/°C

Answer 2

✠ Question Given :

• A copper coil has a resistance of 20.0 ohm at 0 degree celsius & resistance of 26.4 Ohm at 80 degree celsius . Find the temperature Cofficent of resistance of copper.

✠ Required Solution :

✩ Values given to us :

• ⇒ copper coil resistance (Ω) = 20.0 Ω
• ⇒ [ At 0 ° degree Celsius ]

• ⇒ Copper resistance (Ω) = 26.4 Ω
• ⇒ [ At 80° degree Celsius ]

✠ Assumption :

• ⇒ Let β be temperature Cofficent of the wire

✩ Formula used here :

• ⇒ β = R2 - R1 / R1(t2 - t1)

✩ Putting values required :

• ⇒ R1 = 20 Ω

• ⇒ R2 = 26.4 Ω

• ⇒ t1 = 0°C

• ⇒ t2 = 80° C

✩ Solving Further :

• ⇒ β = 26.4 - 20.0 ohm/ 20 ( 80 - 0)

• ⇒ β = 6.4 /20 ( 80 )

• ⇒ β = 6.4 / 1600

• ⇒ β = 4 × 10 -³/°C

✯Temperatures Cofficent is 4 × 10 -³/°C✯

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