A standard coil marked `5 Omega` is found to have a resistance of `5.128 Omega` at `30^@C`. Calculate the temperature at which the marking is correct. The temperature coefficient of resistance of the material of the coil is `.0042^@C^(-1)`.

A standard coil marked `5 Omega` is found to have a resistance of `5.1

1.	A standard coil marked `5 Omega` is found to have a resistance of `5.128 Omega` at `30^@C`. Calculate the temperature at which the marking is correct. The temperature coefficient of resistance of the material of the coil is `.0042^@C^(-1)`.
Answer» `R_(t_1)=5Omega, R_(t_2)5.128 Omega`, `t_(1)=?, t_(2)=30^@C`. `R_(t_1)/R_(t_2)=(R_(0)(1+alpha t_(1)))/(R_(0)(1+alpha t_(2)))=(1 +alpha t_(1))/(1+alpha t_(2))` or `5/5.128= (1+0.0042 xx t_(1))/(1+0.0042 xx30)` On solving, `t_(1)=23.3^@C`

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A standard coil marked `5 Omega` is found to have a resistance of `5.128 Omega` at `30^@C`. Calculate the temperature at which the marking is correct. The temperature coefficient of resistance of the material of the coil is `.0042^@C^(-1)`.

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