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Question Number 36529 by rahul 19 last updated on 03/Jun/18
$$\mathrm{A}\:\mathrm{fuse}\:\mathrm{with}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{cross}−\mathrm{sectional} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{0}.\mathrm{15}\:\mathrm{mm}\:\mathrm{blows}\:\mathrm{at}\:\mathrm{15A}.\:\mathrm{What} \\ $$$$\mathrm{should}\:\mathrm{be}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cross}\:\mathrm{section} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{fuse}\:\mathrm{made}\:\mathrm{of}\:\mathrm{same}\:\mathrm{material}\:\mathrm{that} \\ $$$$\mathrm{blows}\:\mathrm{at}\:\mathrm{30A}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
$${More}\:{area}\:{more}\:{current}\:{required} \\ $$$$\frac{\Pi{r}_{\mathrm{1}} ^{\mathrm{2}} }{\Pi{r}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{{I}_{\mathrm{1}} }{{I}_{\mathrm{2}} } \\ $$$${r}_{\mathrm{2}} ^{\mathrm{2}} =\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} ×{I}_{\mathrm{2}} }{{I}_{\mathrm{1}} }=\frac{\mathrm{0}.\mathrm{15}×\mathrm{0}.\mathrm{15}×\mathrm{30}}{\mathrm{15}} \\ $$$${r}_{\mathrm{2}} =\mathrm{0}.\mathrm{15}×\sqrt{\mathrm{2}}\: \\ $$
Commented by rahul 19 last updated on 03/Jun/18
Thank you sir ����
Answered by math1967 last updated on 03/Jun/18
$${radius}=\sqrt{\frac{.\mathrm{15}×.\mathrm{15}×\mathrm{30}}{\mathrm{15}}\:}=.\mathrm{15}×\sqrt{\mathrm{2}\:}\:{mm} \\ $$$${pls}.\:{check}\:{ans} \\ $$
Commented by rahul 19 last updated on 03/Jun/18
$$\mathrm{Correct}! \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$