Question 344: Current Electricity

Question

The resistance of a wire is $$5$$ ohm at $${50^ \circ }C$$ and $$6$$ ohm at $${100^ \circ }C.$$ The resistance of the wire at $${0^ \circ }C$$ will be

Accepted Answer

KEY CONCEPT :
We know that
$${R_t} = R{}_0\left( {1 + \alpha t} ight),$$
where $${R_t}$$ is the resistance of the wire at $${t^ \circ }C,$$
$${R_0}$$ is the resistance of the wire at $${0^ \circ }C$$
and $$\alpha $$ is the temperature coefficient of resistance
$$ \Rightarrow {R_{50}} = {R_0}\left( {1 + 50\alpha } ight)\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)$$
$${R_{100}} = {R_0}\left( {1 + 100\alpha } ight)\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} ight)$$
From $$\left( i ight),\,{R_{50}} - {R_0} = 50\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} ight)$$
From $$\left( {ii} ight),{R_{100}} - {R_0} = 100\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iv} ight)$$
Dividing $$(iii)$$ by $$(iv),$$ we get
$${{{R_{50}} - {R_0}} \over {{R_{100}} - R{}_0}} = {1 \over 2}$$
Here, $${{R_{50}}}$$ $$ = 5\Omega $$ and $${{R_{100}} = 6\Omega }$$
$$	herefore$$ $${{5 - {R_0}} \over {6 - {R_0}}} = {1 \over 2}$$
or, $$6 - {R_0} = 10 - 2{R_0}$$
or, $${R_0} = 4\Omega .$$

Answer

$$3$$ ohm

Answer

$$2$$ ohm

Answer

$$1$$ ohm

Answer

$$4$$ ohm

Current Electricity

The resistance of a wire is $5 ohm at {50^ \circ }C and 6 ohm at {100^ \circ }C. The resistance of the wire at {0^ \circ }C$ will be

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