Question

When subjected to a voltage of $10 \,V$, the current through a resistor at a temperature of $40^{\circ} C$ is $0.1 \,A$. The temperature coefficient of resistance of the material of the resistor is $2 \times 10^{-4 \circ} C ^{-1}$. The temperature of the resistor in ${ }^{\circ} C$ when the current drops to $0.098 \,A$ is

• A. 142
• B. 167
• C. 181
• D. 206

Answer 1

Answer: (A)

Given,
voltage, $V_{1}=10 V$,
current through resistor, $I_{1}=0.1 A$,
temperature, $t_{1}=40^{\circ} C$
and temperature coefficient, $\alpha=2 \times 10^{-4 \circ} C ^{-1}$
$\therefore $ Resistance of resistor at $t_{1}=40^{\circ} C$
$R_{t_{1}}=\frac{V_{1}}{I_{1}}=\frac{10}{0.1}=100 \Omega$
Since, $R_{t} =R_{0}( l +\alpha t) $
$R_{t_{1}} =R_{0}\left( l +\alpha t_{1}\right) $
$100 =R_{0}\left( l +2 \times 10^{-4} \times 40\right) $
$ R_{0} =\frac{100}{1.008}=99.21 \,\Omega$
At temperature $t_{2}$, current drops to $0.098 A$, then resistance of resistor is given by
$R_{t_{2}}=\frac{V}{I}=\frac{10}{0.098}=102.04 \,\Omega$
Since, $R_{t_{2}}= R_{0}\left(1+\alpha t_{2}\right) $
$102.04= 99.21\left(1+2 \times 10^{-4} t_{2}\right) $
$\Rightarrow 1+2 \times 10^{-4} t_{2} =1.0285 \Rightarrow t_{2}=\frac{0.0285}{2 \times 10^{-4}} $
$=142.6 \approx 142{ }^{\circ} C $