Question
Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, $+ \alpha$, while the other one has a negative temperature of coefficient, $- \alpha$, as shown in the figure. The current through these circuits is denoted by $I_{A}$ and $I_{B}$. At initial temperature, the resistance of the two resistors is $R_{0}$. As the temperature is increased, the correct option that describes the variation of current in these circuits is:
Options
Solution
Sol. [Key Concept: Resistors in series vs parallel]
Figure A
$$R_{\text{eq~}} = R_{0}(1 - \alpha\Delta T + )R_{0}(1 + \alpha\Delta T) = 2R_{0} = Constant$$
$\therefore I_{A} = \frac{V_{}}{2R_{0}}$=Constant
Figure B
$$\begin{matrix} & R_{\text{eq}\text{~}} = \frac{R_{0}(1 + \alpha\Delta T)R_{0}(1 - \alpha\Delta T)}{R_{0}(1 + \alpha\Delta T) + R_{0}(1 - \alpha\Delta T)} \\ & R_{\text{eq}\text{~}} = \frac{R_{0}\left( 1 - \alpha^{2}(\Delta T)^{2} \right)}{2R_{0}} \end{matrix}$$
$R_{\text{eq}\text{~}}$ will decrease with rise in temperature
$$\therefore I_{B} = \frac{V_{0}}{R_{eq}}\text{~}\text{increases with rise in temperature}\text{~}$$
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