Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, + α , while the other one has a negative temperature of coefficient, −

Question

Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, $+ α$ , while the other one has a negative temperature of coefficient, $- α$ , 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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TestHub · Accepted Answer

For circuit (A), the two resistors are connected in series.
The equivalent resistance is given by:
$R_{A} = R_{0} (1 - α Δ T) + R_{0} (1 + α Δ T) = 2 R_{0}$
The current in circuit (A) is $I_{A} = \frac{V}{R _{A}} = \frac{V}{2 R _{0}}$ .
Since $R_{A}$ is independent of the temperature change $Δ T$ , the current $I_{A}$ remains constant.

For circuit (B), the two resistors are connected in parallel.
The equivalent resistance is given by:
$R_{B} = \frac{R _{0} ( 1 - α Δ T ) \cdot R _{0} ( 1 + α Δ T )}{R _{0} ( 1 - α Δ T ) + R _{0} ( 1 + α Δ T )}$
$R_{B} = \frac{R _{0}^{2} ( 1 - α ^{2} Δ T ^{2} )}{2 R _{0}} = \frac{R _{0}}{2} (1 - α^{2} Δ T^{2})$
As the temperature increases, $Δ T$ increases, which means the term $(1 - α^{2} Δ T^{2})$ decreases.
Therefore, the equivalent resistance $R_{B}$ decreases.
The current in circuit (B) is $I_{B} = \frac{V}{R _{B}}$ .
Since $R_{B}$ decreases with an increase in temperature, the current $I_{B}$ increases.

Thus, $I_{A}$ remains constant while $I_{B}$ increases.

Answer: $I_{A}$ remains constant while $I_{B}$ increases

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