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Question Number 36548 by rahul 19 last updated on 03/Jun/18

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

$$Q=\frac{kA({\theta }_2-{\theta }_1)t}{l}=\frac{i^{2}Rt}{J}=\frac{{\left(V\right)}^{2}t}{RJ}$$$$A={(40-2\times 0.1)}^2=1584.04{cm}^2$$$${\theta }_2-{\theta }_1=100$$$$l=0.1cm$$$$V=100Volt$$$$J=$$$$R=\frac{{\left(V\right)}^{2}l}{JkA({\theta }_2-{\theta }_1)}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

$$heattransfer...attachingtheorem...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

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