The V – I graph for a conductor at temperature $T_{1} and T_{2}$ are as shown in figure. Then ( $T_{2} - T_{1}$ ) is equal to [ $α$ = temperature coefficient of resistivity]

Question

The V – I graph for a conductor at temperature $$T1$$ and $$T2$$ are as shown in figure. Then (T2$$– $$T1) is equal to [α $$=$$ temperature coefficient of resistivity]

Infinity Learn · Accepted Answer

According to Ohm’s Law, V $$=$$ R.I. It means that ‘R’ is the slope of V – I graph.∴   $$R1=$$$$tan$$θ  and ∴   $$R2=R1$$  $$1+$$α $$T2$$−$$T1$$ ∴  $$cot$$θ  $$=$$  $$tan$$θ  $$1+$$α $$T2$$−$$T1$$ ⇒  α  $$T2$$−$$T1=$$$$cot$$θ$$tan$$θ−$$1$$  ⇒  $$T2$$−$$T1=1$$α .  $$cos2$$θ$$sin2$$θ $$=$$ $$1$$α  $$cos600sin2300=2$$α

The V – I graph for a conductor at temperature $T_{1} and T_{2}$ are as shown in figure. Then ( $T_{2} - T_{1}$ ) is equal to [ $α$ = temperature coefficient of resistivity]

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The V – I graph for a conductor at temperature T1 and T2 are as shown in figure. Then (T2– T1) is equal to [α = temperature coefficient of resistivity]

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The V – I graph for a conductor at temperature $T_{1} and T_{2}$ are as shown in figure. Then ( $T_{2} - T_{1}$ ) is equal to [ $α$ = temperature coefficient of resistivity]