Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are

Q: Why can we just ignore the $\alpha_1 \alpha_2$ term in the parallel derivation?

Temperature coefficients for metals are typically very small, around $10^{-3}$ to $10^{-5} \, ^{\circ}\text{C}^{-1}$. When you multiply two such small numbers, the result is on the order of $10^{-6}$ to $10^{-10}$, which is so tiny compared to the other terms that it has virtually no impact on the final result.

Q: I don't understand how the binomial approximation was applied.

We had a term like $\frac{1}{1 + x}$. Using the binomial theorem, this can be written as $(1 + x)^{-1}$. For very small values of $x$, the higher-order terms in the expansion become negligible, leaving us with just $1 - x$. This is a standard trick in physics to simplify complex fractions!

Q: Is the effective temperature coefficient always the average of the two?

In this specific case, yes, because both conductors had the exact same initial resistance $R_0$. If their initial resistances were different, the effective coefficient would be a weighted average, leaning closer to the coefficient of the conductor with the higher resistance.

Q: Why did we drop the $\theta^2$ term after multiplying the brackets?

Just like the $\alpha_1 \alpha_2$ term, the $\theta^2$ term is multiplied by $\alpha^2$. Since $\alpha$ is already very small, $\alpha^2$ is microscopically small. In linear approximations, we only keep terms up to the first power of the small variable.

Q: Does this logic apply to semiconductors as well?

Semiconductors have negative temperature coefficients (their resistance drops as they heat up). However, as long as the temperature change isn't drastic and the linear approximation $R = R_0(1 + \alpha \theta)$ holds true, the algebraic derivation remains perfectly valid.

Question

Two conductors have the same resistance at $0^{\circ}	ext{C}$ but their temperature coefficients of resistance are $\alpha_1$ and $\alpha_2$. The respective temperature coefficients of their series and parallel combinations are nearly

Options:

(A) $\frac{\alpha_1 + \alpha_2}{2}, \alpha_1 + \alpha_2$

(B) $\alpha_1 + \alpha_2, \frac{\alpha_1 + \alpha_2}{2}$

(C) $\alpha_1 + \alpha_2, \frac{\alpha_1 \alpha_2}{\alpha_1 + \alpha_2}$

(D) $\frac{\alpha_1 + \alpha_2}{2}, \frac{\alpha_1 + \alpha_2}{2}$

Sigma Percentile · Accepted Answer

### The Magic of Temperature Coefficients

Imagine you have two identical-looking wires. At the freezing point of water, $0^{\circ}	ext{C}$, they both offer the exact same resistance to the flow of electricity, let's call it $R_0$.

But here is where the plot thickens. As things heat up, these wires reveal their true colors. Their resistances increase, but at different rates! This rate of change is governed by their temperature coefficients of resistance, $\alpha_1$ and $\alpha_2$.

The burning question is: what happens when we team them up? What is the effective temperature coefficient if we connect them in series, or in parallel? Let's dive into the math and uncover the hidden symmetry.

### The Series Connection: A Straightforward Addition

When we connect resistors in series, their resistances simply add up.

$$R_s = R_1 + R_2$$

We know how individual resistances behave with temperature $	heta$:

$$R_1 = R_0(1 + \alpha_1 	heta)$$
$$R_2 = R_0(1 + \alpha_2 	heta)$$

Let's plug these into our series equation. The equivalent resistance at $0^{\circ}	ext{C}$ is just $R_{s0} = R_0 + R_0 = 2R_0$. So, the series combination's resistance at temperature $	heta$ is:

$$2R_0(1 + \alpha_s 	heta) = R_0(1 + \alpha_1 	heta) + R_0(1 + \alpha_2 	heta)$$

Notice how $R_0$ is everywhere? Let's divide the entire equation by $R_0$ to clean things up.

$$2(1 + \alpha_s 	heta) = (1 + \alpha_1 	heta) + (1 + \alpha_2 	heta)$$

$$2 + 2\alpha_s 	heta = 2 + (\alpha_1 + \alpha_2)	heta$$

Subtracting 2 from both sides and dividing by $2	heta$, we arrive at a beautifully simple result:

$$\alpha_s = \frac{\alpha_1 + \alpha_2}{2}$$

The effective temperature coefficient in series is just the arithmetic mean of the individual coefficients!

### The Parallel Connection: A Test of Approximations

Now, let's tackle the parallel combination. The formula here is a bit more intimidating:

$$R_p = \frac{R_1 R_2}{R_1 + R_2}$$

At $0^{\circ}	ext{C}$, the equivalent resistance is $R_{p0} = \frac{R_0 \cdot R_0}{R_0 + R_0} = \frac{R_0}{2}$. Let's substitute our temperature-dependent expressions:

$$\frac{R_0}{2}(1 + \alpha_p 	heta) = \frac{R_0(1 + \alpha_1 	heta) \cdot R_0(1 + \alpha_2 	heta)}{R_0(1 + \alpha_1 	heta) + R_0(1 + \alpha_2 	heta)}$$

Let's expand the numerator. We get $1 + \alpha_1 	heta + \alpha_2 	heta + \alpha_1 \alpha_2 	heta^2$.

**Here is the crucial physical insight:** Temperature coefficients ($\alpha$) for metals are incredibly small, typically on the order of $10^{-3}$ to $10^{-5}$. When you multiply two such tiny numbers together ($\alpha_1 \alpha_2$), the result is microscopic! We can safely ignore the $\alpha_1 \alpha_2 	heta^2$ term.

The denominator simplifies to $2 + (\alpha_1 + \alpha_2)	heta$. Canceling one $R_0$ from the top and bottom, our equation becomes:

$$\frac{1}{2}(1 + \alpha_p 	heta) = \frac{1 + (\alpha_1 + \alpha_2)	heta}{2 + (\alpha_1 + \alpha_2)	heta}$$

Let's multiply both sides by 2 to get rid of that fraction on the left:

$$1 + \alpha_p 	heta = \frac{1 + (\alpha_1 + \alpha_2)	heta}{1 + \frac{\alpha_1 + \alpha_2}{2}	heta}$$

### The Binomial Lifesaver

We have a temperature-dependent term in the denominator. To solve this elegantly, we use the **Binomial Approximation**: $(1 + x)^{-1} \approx 1 - x$ for very small values of $x$.

Applying this to our denominator:

$$1 + \alpha_p 	heta \approx (1 + (\alpha_1 + \alpha_2)	heta) \left(1 - \frac{\alpha_1 + \alpha_2}{2}	hetaight)$$

Now, let's multiply these brackets. Once again, we will encounter a $	heta^2$ term, and once again, we will ruthlessly ignore it because it's negligibly small.

$$1 + \alpha_p 	heta \approx 1 + (\alpha_1 + \alpha_2)	heta - \frac{\alpha_1 + \alpha_2}{2}	heta$$

$$1 + \alpha_p 	heta \approx 1 + \left(\alpha_1 + \alpha_2 - \frac{\alpha_1 + \alpha_2}{2}ight)	heta$$

$$1 + \alpha_p 	heta \approx 1 + \left(\frac{\alpha_1 + \alpha_2}{2}ight)	heta$$

Comparing both sides, we reach our grand conclusion:

$$\alpha_p = \frac{\alpha_1 + \alpha_2}{2}$$

### The Grand Symmetry

Isn't nature poetic? Whether you connect these identical-resistance conductors in series or in parallel, the effective temperature coefficient is exactly the same! It is always the arithmetic mean of their individual coefficients.

This problem is a masterclass in how physicists use mathematical approximations to cut through algebraic clutter and reveal the elegant truths hidden underneath.

Animated Solution for Physics - Current Electricity: Two conductors have the same resistance at $0^{\circ} C$ but their temperature coefficients of resistance are $α_{1}$ and $α_{2}$ . The respective temperature coefficients of their series and parallel combinations are nearly

Select Answer:

Visualized Solution

$R_{1}$ and $R_{2}$ at temperature $θ$

Series Combination Setup

Substituting Values in Series

Solving for $α_{s}$

Parallel Combination Setup

Substituting Values in Parallel

Expanding Numerator and Denominator

Binomial Approximation

Solving for $α_{p}$

Final Conclusion

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Magic of Temperature Coefficients

The Series Connection

The Parallel Connection

The Binomial Lifesaver

The Grand Symmetry

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