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 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:

• A. Both \(I_A\) and \(I_B\) remain constant
• B. \(I_A\) remains constant while \(I_B\) increases
• C. \(I_A\) decreases while \(I_B\) increases
• D. \(I_A\) increases while \(I_B\) decreases

Hint:

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.

Answer 1

The Correct Option is B

Concept:

• Resistance at temperature change \(\Delta T\) is given by \[ R=R_0(1+\alpha\Delta T) \] for positive temperature coefficient.

• For a negative temperature coefficient, \[ R=R_0(1-\alpha\Delta T) \]

• In a series combination, equivalent resistance is the sum of individual resistances.

• In a parallel combination, \[ R_{\text{eq}}=\frac{R_1R_2}{R_1+R_2} \]

• Current supplied by a battery is \[ I=\frac{V}{R_{\text{eq}}} \] Hence the variation of current depends upon the variation of equivalent resistance.

Step 1: Find equivalent resistance of circuit (A)
The two resistors are connected in series. \[ R_1=R_0(1-\alpha\Delta T) \] \[ R_2=R_0(1+\alpha\Delta T) \] Therefore, \[ R_A=R_1+R_2 \] \[ R_A=R_0(1-\alpha\Delta T)+R_0(1+\alpha\Delta T) \] \[ R_A=R_0\left[(1-\alpha\Delta T)+(1+\alpha\Delta T)\right] \] \[ R_A=R_0(2) \] \[ R_A=2R_0 \] Thus, the equivalent resistance of circuit (A) is independent of temperature.

Step 2: Determine current in circuit (A)
Using Ohm's law, \[ I_A=\frac{V}{R_A} \] \[ I_A=\frac{V}{2R_0} \] Since \(R_A\) does not change with temperature, \[ I_A=\text{constant} \] Hence the current in circuit (A) remains unchanged.

Step 3: Find equivalent resistance of circuit (B)
The two resistors are connected in parallel. \[ R_B=\frac{R_1R_2}{R_1+R_2} \] Substituting the values, \[ R_B= \frac{R_0(1-\alpha\Delta T)\,R_0(1+\alpha\Delta T)} {R_0(1-\alpha\Delta T)+R_0(1+\alpha\Delta T)} \] \[ R_B= \frac{R_0^2\left(1-\alpha^2\Delta T^2\right)} {2R_0} \] \[ R_B= \frac{R_0}{2} \left(1-\alpha^2\Delta T^2\right) \]

Step 4: Study the variation of equivalent resistance
Since \[ \alpha^2\Delta T^2\gt 0 \] it follows that \[ 1-\alpha^2\Delta T^2\lt 1 \] Therefore, \[ R_B\lt \frac{R_0}{2} \] Thus the equivalent resistance decreases as temperature increases.

Step 5: Determine current in circuit (B)
Using Ohm's law, \[ I_B=\frac{V}{R_B} \] As \(R_B\) decreases, \[ I_B \] must increase. Hence, \[ I_B \text{ increases with temperature.} \]

Step 6: Choose the correct option
We have obtained: \[ I_A=\text{constant} \] and \[ I_B=\text{increasing} \] Therefore the correct statement is \[ \boxed{\text{\(I_A\) remains constant while \(I_B\) increases}} \]