Animated Solution for Physics - Current Electricity: The energy dissipated by a resistor is $10\text{ mJ}$ in $1\text{ s}$ when an electric current of $2\text{ mA}$ flows through it. The resistance is ......... $\Omega$. (Round off to the nearest integer)

The phenomenon of electrical heating is something we experience every single day. From the warm glow of an incandescent bulb to the comforting heat of a toaster, the conversion of electrical energy into thermal energy is a cornerstone of modern technology. But how exactly does this happen? And more importantly, how can we quantify it?

In this problem, we are given a scenario where a resistor dissipates a specific amount of energy as heat when a known current flows through it for a given time. Our mission is to uncover the hidden property of this component: its electrical resistance.

Imagine a microscopic journey inside the conducting material of a resistor. When a voltage is applied, an electric field is established, urging the free electrons to move. As these electrons accelerate, they don't get a clear, unobstructed path. Instead, they constantly collide with the vibrating, positively charged ions of the crystal lattice.

With every collision, the electrons lose some of their kinetic energy, transferring it to the lattice ions. This transferred energy causes the ions to vibrate even more vigorously. On a macroscopic scale, this increased microscopic vibration manifests as an increase in temperature. The resistor heats up, and this thermal energy is dissipated into the surroundings.

This beautiful relationship between electricity and heat was extensively studied by James Prescott Joule in the 19th century, leading to what we now call Joule's Law of Heating.

Joule's Law of Heating provides a precise mathematical formulation for this energy conversion. It states that the heat energy ($H$) dissipated by a resistor is directly proportional to the square of the electric current ($I$) flowing through it, the resistance ($R$) of the conductor, and the time ($t$) for which the current flows.

Mathematically, this is expressed as:

In our problem, we are given the heat dissipated, the current, and the time. We need to find the resistance. By simple algebraic rearrangement, we can isolate $R$:

This is our master equation for the problem. It tells us exactly what operations we need to perform to find the answer.

Unit Conversion

Before we rush into plugging numbers into our equation, we must pause and inspect the units. Physics is unforgiving when it comes to mismatched units. To ensure our final answer for resistance comes out in standard Ohms ($\Omega$), all our input variables must be in their respective standard SI units: Joules for energy, Amperes for current, and seconds for time.

Let's look at what we are given: 1. Energy ($H$): $10\text{ mJ}$ (millijoules). The prefix "milli" means $10^{-3}$. So, we convert this to standard Joules:

2. Current ($I$): $2\text{ mA}$ (milliamperes). Again, applying the "milli" prefix:

3. Time ($t$): $1\text{ s}$. This is already in the standard SI unit, so no conversion is needed.

Now that our units are perfectly aligned, we can substitute these values into our rearranged master equation.

Let's break down the calculation step-by-step to avoid any silly mistakes. First, we need to handle the denominator, specifically the square of the current.

When squaring a term in scientific notation, we square the coefficient and multiply the exponent by 2:

Substituting this back into our equation, we get:

Now, we can separate the coefficients from the powers of 10 to make the division easier:

Dividing the coefficients is straightforward:

For the powers of 10, we use the rule of exponents which states that when dividing terms with the same base, we subtract the exponent of the denominator from the exponent of the numerator:

Combining these results, we arrive at our final expression for the resistance:

To express the answer as a standard integer, we simply multiply $2.5$ by $10^3$ (which is $1000$):

The resistance of the component is exactly $2500\Omega$.

This problem serves as a fantastic reminder of two core principles in physics problem-solving: always start with the fundamental governing law (Joule's Law in this case), and never underestimate the importance of strict unit conversions before performing calculations.