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 coefficient of resistance, \(-\alpha\), as shown in the figure.
The current through these circuits are 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:
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In series combination, directly add resistances and check whether temperature-dependent terms cancel.
For parallel combinations, use the product-over-sum formula carefully.
If equivalent resistance decreases, current increases for a fixed voltage source.
A resistor with positive temperature coefficient increases in resistance, while a negative coefficient decreases in resistance.
Step 1: Set up the two resistors. Each resistor starts at $R_0$. One has coefficient $+\alpha$ so it becomes $R_0(1+\alpha\Delta T)$; the other has $-\alpha$ so it becomes $R_0(1-\alpha\Delta T)$ as temperature rises by $\Delta T$. Step 2: Circuit (A) is a series pair. Add the two resistances. \[ R_A = R_0(1+\alpha\Delta T) + R_0(1-\alpha\Delta T) = 2R_0 \] Step 3: Current in (A). Since $R_A = 2R_0$ has no temperature dependence, $I_A = V/(2R_0)$ stays constant. Step 4: Circuit (B) is a parallel pair. Use $R_B = R_1 R_2/(R_1+R_2)$. The denominator is again $2R_0$, and the numerator is $R_0^2(1-\alpha^2\Delta T^2)$. \[ R_B = \frac{R_0}{2}\left(1 - \alpha^2 \Delta T^2\right) \] Step 5: How $R_B$ behaves. Because $\alpha^2 \Delta T^2 > 0$, the bracket is less than $1$, so $R_B$ drops below $R_0/2$ as temperature rises. Step 6: Current in (B) and conclusion. With $I_B = V/R_B$ and $R_B$ falling, $I_B$ increases. So $I_A$ stays constant while $I_B$ increases. \[ \boxed{I_A \text{ remains constant while } I_B \text{ increases}} \]
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