Q.A heating element using nichrome connected to a $230\ \text{V}$ supply draws an initial current of $3.2\ \text{A}$ which settles after a few seconds to a steady value of $2.8\ \text{A}$. What is the steady temperature of the heating element if the room temperature is $27.0\ ^\circ\text{C}$? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70 \times 10^{-4}\ ^\circ\text{C}^{-1}$.

Concept: Temperature Dependence of Resistance — as temperature rises, resistance increases, reducing the current at constant voltage.

Reasoning:

1. From Ohm’s law, initial resistance at room temperature $R_0 = \frac{230}{3.2} = 71.875\ \Omega$.
2. Steady-state resistance $R_t = \frac{230}{2.8} \approx 82.143\ \Omega$.
3. Using $R_t = R_0 [1 + \alpha (T - T_0)]$ with $\alpha = 1.70 \times 10^{-4}\ ^\circ\text{C}^{-1}$ and $T_0 = 27.0\ ^\circ\text{C}$: …

The key idea here is that the heating element's resistance increases with temperature. When you first switch it on, it's cold — so it draws more current. As it heats up, resistance rises, current drops, and eventually it reaches a steady thermal equilibrium where the electrical power dissipated equals the heat lost to the surroundings. The problem gives us the current at both the cold (initial) and hot (steady) states, and the supply voltage is fixed at $230\ \text{V}$. That means we can compute the resistance at each state using Ohm's law, and then use the temperature coefficient formula to find the temperature rise.

Let's walk through it.

1. Find the initial (cold) resistance. At room temperature $T_0 = 27.0\ ^\circ\text{C}$, the initial current is $I_0 = 3.2\ \text{A}$. Using Ohm's law:

2. Find the steady (hot) resistance. After the element heats up, the current settles to $I = 2.8\ \text{A}$.

We'll keep it as $82.143\ \Omega$ for calculation.

3. Recall the temperature dependence of resistance. For most metals (nichrome is a nickel-chromium alloy, a metal), resistance increases approximately linearly with temperature over a moderate range:

where $\alpha$ is the temperature coefficient of resistance, $T_0$ is the reference temperature (here room temperature), and $T$ is the final temperature.

4. Rearrange to solve for $T$.

Then:

5. Plug in the numbers.

So: …

Method: Using the Temperature Coefficient of Resistance Formula

This method relates the change in resistance of a conductor to the change in temperature using the temperature coefficient of resistance ($\alpha$).

Step-by-step solution

Step 1: Find initial resistance at room temperature

Using Ohm’s law at the moment the circuit is switched on (when element is still at room temperature $27.0\ ^\circ\text{C}$):

Step 2: Find steady-state resistance at higher temperature

Using Ohm’s law after the current stabilises:

Step 3: Apply the temperature dependence formula

The relation between resistance and temperature (for small $\alpha$ and moderate temperature ranges) is:

Where:

• $R_t$ = resistance at temperature $T$ (steady value)
• $R_0$ = resistance at reference temperature $T_0$ (room temperature) …

Here are the common mistakes students make on this problem and how to avoid each.

1. Confusing Initial and Steady Conditions

Mistake:

Using the initial current ($3.2\ \text{A}$) to calculate the hot resistance, or using the steady current ($2.8\ \text{A}$) to calculate the cold resistance.

Why it’s wrong:

• The initial current flows when the element is cold (room temperature).
• The steady current flows when the element has heated up to its final temperature.

How to avoid:

Clearly label:

• Cold state: $I_0 = 3.2\ \text{A}$, $T_0 = 27.0\ ^\circ\text{C}$
• Hot state: $I = 2.8\ \text{A}$, $T = ?$

Then use Ohm’s law:

• Cold resistance: $R_0 = \frac{V}{I_0}$
• Hot resistance: $R = \frac{V}{I}$

2. Forgetting to Use the Correct Formula for Temperature Dependence

Mistake:

Using $R = R_0 (1 + \alpha \Delta T)$ but plugging in the wrong resistance values or forgetting that $\Delta T = T - T_0$.

How to avoid:

Write the formula explicitly:

Then substitute $R$ and $R_0$ from Ohm’s law:

Cancel $V$ (common mistake: forgetting to cancel — leads to extra work and errors).

3. Incorrectly Solving for $T$

Mistake:

Solving algebraically but making sign errors or forgetting to add $T_0$ at the end.

Common slip:

After finding $\Delta T$, some students write $T = \Delta T$ instead of $T = T_0 + \Delta T$.

How to avoid:

Follow step-by-step:

1. Cancel $V$:

2. Rearrange:

3. Solve for $T - T_0$:

4. Finally:

Always write the final step explicitly.

4. Arithmetic Errors with Powers of Ten

Mistake:

Mishandling $\alpha = 1.70 \times 10^{-4}$ — either misplacing the decimal or dividing incorrectly.

Example error:

Computing $\frac{1}{\alpha} \approx 5882$ but then using $5882 \times 0.1429$ incorrectly.

How to avoid:

• Use fractions or keep scientific notation until the final step.
• Double-check: $\frac{1}{\alpha} = \frac{1}{1.70 \times 10^{-4}} = \frac{10^4}{1.70} \approx 5882.35$ …