Animated Solution for Physics - Current Electricity: The resistance of a wire is $5\Omega$ at $50^{\circ }\text{C}$ and $6\Omega$ at $100^{\circ }\text{C}$. The resistance of the wire at $0^{\circ }\text{C}$ will be

Unraveling the Temperature Dependence of Resistance

Have you ever noticed how your laptop gets warm when you run heavy applications? Or how a light bulb glows brightly when electricity passes through it? This happens because of resistance, and interestingly, resistance itself isn't a constant number—it changes with temperature!

In this problem, we are given a wire whose resistance is $5\Omega$ at $50^{\circ }\text{C}$ and $6\Omega$ at $100^{\circ }\text{C}$. Our mission is to find its resistance at the freezing point of water, $0^{\circ }\text{C}$. Let's dive into the math behind this physical phenomenon.

For metallic conductors, the resistance increases linearly with temperature over a moderate range. The relationship is beautifully captured by the formula:

Here, $R_t$ is the resistance at temperature $t$, $R_0$ is the resistance at $0^{\circ }\text{C}$, and $\alpha$ is the temperature coefficient of resistance. We have two unknowns: $R_0$ and $\alpha$. Thankfully, the problem gives us two data points, which means we can set up a system of two linear equations.

Let's plug in our first data point. At $t=50^{\circ }\text{C}$, the resistance is $5\Omega$:

Now, let's plug in the second data point. At $t=100^{\circ }\text{C}$, the resistance is $6\Omega$:

We have a classic algebra puzzle. The most elegant way to solve this is to divide equation (i) by equation (ii). Why? Because it instantly eliminates $R_0$, leaving us with an equation containing only $\alpha$:

Now, we cross-multiply to solve for $\alpha$:

We've found the temperature coefficient! Now, we just need to substitute this value back into our first equation to find $R_0$:

And there we have it! The resistance of the wire at $0^{\circ }\text{C}$ is exactly $4\Omega$. This linear relationship is a fundamental concept in current electricity, and mastering it allows you to predict how circuits will behave under varying thermal conditions.