Animated Solution for Physics - Current Electricity: A current of $2\text{ mA}$ was passed through an unknown resistor which dissipated a power of $4.4\text{ W}$. Dissipated power when an ideal power supply of $11\text{ V}$ is connected across it is

The behavior of electrical components can sometimes feel like a puzzle, but the laws of physics always provide the key. In this problem, we are introduced to a mysterious resistor. We don't know its resistance, but we do know how it behaves under a specific condition: when a tiny current of $2\text{ mA}$ flows through it, it dissipates $4.4\text{ W}$ of power as heat.

Our mission is to predict how this exact same resistor will behave when connected to a completely different power source—an ideal $11\text{ V}$ supply. To do this, we must first uncover the intrinsic property of the resistor: its resistance $R$.

Power dissipated by a resistor can be expressed in multiple ways, depending on the known variables. Since we are given the current $I$ and the power $P$, the most direct formula to use is:

Before we plug in the numbers, we must ensure all our units are in the standard SI format. The current is given in milliamperes, so we convert it to amperes:

Now, let's substitute our known values into the power equation:

Squaring the current gives us $4\times 10^{-6}$. Now, we can isolate $R$:

Dividing $4.4$ by $4$ gives $1.1$, and bringing the power of ten to the numerator gives us:

We have successfully unmasked the resistor! It has a massive resistance of $1.1\text{ M}\Omega$. Because resistance is an intrinsic property of the component (assuming temperature remains constant), this value will not change when we move the resistor to a new circuit.

Now, we take our $1.1\text{ M}\Omega$ resistor and connect it across an ideal $11\text{ V}$ power supply. An "ideal" supply means it has zero internal resistance, so the full $11\text{ V}$ is applied directly across our resistor.

We need to find the new power dissipation, $P_2$. This time, we know the voltage $V$ and the resistance $R$. The most efficient power formula to use here is:

Let's substitute our values into this equation:

We know that $11^2=121$. Substituting this in, we get:

Dividing $121$ by $1.1$ gives exactly $110$. So, the expression becomes:

To match the format of the given options, we can rewrite this by shifting the decimal point one place to the left, which increases the exponent by one:

Conclusion: By first using the current and power to find the resistance, and then using that resistance with the new voltage, we seamlessly predicted the resistor's behavior in a new environment. The power dissipated in the second scenario is $11\times 10^{-5}\text{ W}$, which corresponds perfectly to option (b).