Question

When subjected to a voltage of $10V$, the current through a resistor at a temperature of $40^{\circ }C$ is $0.1A$. 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 $0.098A$ is

• A. 142
• B. 167
• C. 181
• D. 206

Answer 1

Answer: (A)

Given,
voltage, $V_1=10V$,
current through resistor, $I_1=0.1A$,
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 $0.098A$, then resistance of resistor is given by
$R_{t_2}=\frac{V}{I}=\frac{10}{0.098}=102.04\Omega$
Since, $R_{t_2}=R_0\left(1+\alpha t_2\right)$
$102.04=99.21\left(1+2\times 10^{-4}{t}_2\right)$
$\Rightarrow 1+2\times 10^{-4}{t}_2=1.0285\Rightarrow t_2=\frac{0.0285}{2\times 10^{-4}}$
$=142.6\approx 142{}^{\circ }C$