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  • The resistance of a conductor depends on the tension that the conductor is subjected to through the r

The resistance of a conductor depends on the tension \mathrm{T} that the conductor is subjected to through the relation
\mathrm{\mathrm{R}^{\prime}=\mathrm{R}_{0}(1+\alpha \mathrm{T})}
Suppose, further, that the tension varies as
\mathrm{\mathrm{T}=\mathrm{T}_{0} \sin \omega \mathrm{t}}.
If the conductor is connected in the circuit shown in the figure, The heat dissipated in the circuit is:

Option: 1

\mathrm{\frac{E^{2}}{2 R_{0}} \frac{1}{\sqrt{1-\frac{\alpha^{2} T_{0}^{2}}{4}}}}


Option: 2

\mathrm{\frac{E^{2}}{R_{0}} \frac{1}{\sqrt{1-\frac{\alpha^{2} T_{0}^{2}}{4}}}}


Option: 3

\mathrm{\frac{2 E^{2}}{R_{0}} \frac{1}{\sqrt{1-\frac{\alpha^{2} T_{0}^{2}}{4}}}}


Option: 4

None of the above


Answers (1)

best_answer

The resistance of the conductor is given by : \mathrm{R^{\prime}= R_{0}\left(1+\alpha T_{0} \sin \omega t\right)}
The current is
\mathrm{I=\frac{E}{R_{0}+R^{\prime}}=\frac{E}{R_{0}+R_{0}\left(1+\alpha T_{0} \sin \omega t\right)}}
\mathrm{or, I=\frac{E}{2 R_{0}\left(1+\frac{\alpha T_{0}}{2} \sin \omega t\right)}}

The average heat dissipated in the circuit is given by (over a single cycle):
\mathrm{Q_{a v} =\frac{1}{T} \int_{0}^{2 \pi} \frac{\mathrm{E}^{2}}{\mathrm{R}_{0}+\mathrm{R}^{\prime}} \mathrm{dt} ; \quad \mathrm{T}=\frac{2 \pi}{\omega}}
         \mathrm{=\frac{\mathrm{E}^{2}}{2 \mathrm{R}_{0}} \times \frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{\mathrm{d} \theta}{1+\beta \sin \theta)} ; \quad \text { Where } \beta=\frac{\alpha \mathrm{T}_{0}}{2} \& \theta=\omega \mathrm{t}}
         \mathrm{=\frac{\mathrm{E}^{2}}{2 \mathrm{R}_{0}} \times \frac{1}{\sqrt{1-\beta^{2}}}}
        \mathrm{=\frac{\mathrm{E}^{2}}{2 \mathrm{R}_{0}} \frac{1}{\sqrt{1-\frac{\alpha^{2} \mathrm{~T}_{0}^{2}}{4}}}}  

 

Posted by

seema garhwal

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