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kaheart [24]
3 years ago
11

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

, the initial current decreases until the filament reaches its steady-state temperature. The temperature coefficient of resistivity of the filament is 4 times 10-3 K-1. The final current through the filament is one- eighth the initial current. What is the change in temperature of the filament
Physics
1 answer:
alekssr [168]3 years ago
6 0

Answer:

The change in temperature is \Delta T  = 1795 K

Explanation:

From the question we  are told that

   The temperature coefficient is  \alpha  =  4 * 10^{-3 }\  k^{-1 }

The resistance of the filament is mathematically represented as

           R  =  R_o [1 + \alpha  \Delta T]

Where R_o is the initial resistance

Making the change in temperature the subject of the formula

     \Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]

Now from ohm law

           I = \frac{V}{R}

This implies that current varies inversely with current so

           \frac{R}{R_o} = \frac{I_o}{I}

Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

    I  = \frac{I_o}{8}

Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

        \Delta T  = 1795 K

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To calculate the new angular velocity, we relate it to the conservation of angular momentum of the system I.e when the bug was at the edge of the disk and when it is located at the centre of the disk.

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∇K = ½(I₂*ω₂² - I₁ω₁²)

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