Copper Wire Resistance And Current Calculation

Hey guys! Let's dive into a fun physics problem today. We're going to calculate the resistance and current flowing through a copper wire. This is a classic example that combines the concepts of resistivity, resistance, and Ohm's Law. So, buckle up, and let's get started!

Problem Statement

We have a copper wire that's 2.5 meters long and has a cross-sectional area of 1.0 mm². The resistivity of copper is given as $1.7 imes 10^{-8} \Omega\cdot m$. We need to:

a) Determine the resistance of the wire. b) Calculate the current flowing through the wire when a potential difference of 12 V is applied across it.

Part A: Determining the Resistance of the Wire

To find the resistance of the wire, we'll use the formula that relates resistance to resistivity, length, and cross-sectional area. The formula is:

$R = \rho \frac{L}{A}$

Where:

• $R$ is the resistance (in Ohms, $\Omega$)
• $\rho$ is the resistivity (in Ohm-meters, $\Omega\cdot m$)
• $L$ is the length of the wire (in meters, m)
• $A$ is the cross-sectional area of the wire (in square meters, m²)

Step 1: Convert Units

Before we plug in the values, we need to make sure our units are consistent. The length is already in meters (2.5 m), but the area is given in mm². We need to convert mm² to m². Remember, 1 mm = $10^{-3}$ m, so 1 mm² = $(10^{-3} m)^2 = 10^{-6} m²$.

Therefore, the cross-sectional area, $A = 1.0 \, mm^2 = 1.0 imes 10^{-6} \, m^2$.

Step 2: Plug in the Values

Now we can plug the values into the formula:

$R = (1.7 imes 10^{-8} \Omega\cdot m) \frac{2.5 \, m}{1.0 imes 10^{-6} \, m^2}$

Step 3: Calculate the Resistance

Let's do the math:

$R = (1.7 imes 10^{-8}) imes (2.5 imes 10^{6}) \, \Omega$

$R = 4.25 imes 10^{-2} \, \Omega$

So, the resistance of the copper wire is 0.0425 Ohms. That's a pretty low resistance, which makes sense since copper is a good conductor of electricity.

Key Takeaway: This part highlights the direct relationship between the length of a conductor and its resistance, and the inverse relationship between the cross-sectional area and resistance. Understanding this relationship is crucial for designing electrical circuits and selecting appropriate wires for different applications. The resistivity itself is an intrinsic property of the material, reflecting how well it resists the flow of electric current. Copper, with its low resistivity, is an excellent choice for wiring due to its minimal energy loss during current transmission.

Part B: Calculating the Current

Now that we know the resistance of the wire, we can calculate the current flowing through it when a 12 V potential difference is applied. For this, we'll use Ohm's Law, which is a fundamental principle in electrical circuits.

Ohm's Law states:

$V = IR$

Where:

• $V$ is the potential difference (in Volts, V)
• $I$ is the current (in Amperes, A)
• $R$ is the resistance (in Ohms, $\Omega$)

Step 1: Rearrange Ohm's Law to Solve for Current

We need to find the current ($I$), so we'll rearrange the formula:

$I = \frac{V}{R}$

Step 2: Plug in the Values

We have $V = 12 \, V$ and we calculated $R = 0.0425 \, \Omega$ in the previous part. Let's plug these values into the formula:

$I = \frac{12 \, V}{0.0425 \, \Omega}$

Step 3: Calculate the Current

Now, let's do the division:

$I \approx 282.35 \, A$

So, the current flowing through the copper wire is approximately 282.35 Amperes. That's a significant amount of current! This illustrates how a small resistance can lead to a large current when a potential difference is applied.

Key Takeaway: This high current value underscores the importance of understanding the interplay between voltage, current, and resistance in electrical circuits. In practical applications, such high currents can generate significant heat due to the power dissipated in the wire ($P = I^2R$). This heat can potentially damage the wire or surrounding components, highlighting the need for appropriate circuit protection measures, such as fuses or circuit breakers, to prevent overheating and potential hazards.

Conclusion

We've successfully calculated the resistance and current in our copper wire example. We found that the resistance was 0.0425 Ohms, and when a 12 V potential difference was applied, the current was approximately 282.35 Amperes.

This exercise demonstrates the practical application of fundamental electrical concepts like resistivity, resistance, and Ohm's Law. By understanding these concepts, we can analyze and design electrical circuits effectively. Remember guys, always double-check your units and pay attention to the magnitude of your results. A result like 282.35 Amperes should immediately raise a flag and prompt you to consider the implications and potential dangers of such a high current in a real-world scenario. Always stay safe and keep learning!

Final Thoughts: These calculations are crucial for anyone working with electrical systems, from designing simple circuits to understanding power distribution networks. The ability to determine resistance and current flow allows engineers and technicians to ensure the safe and efficient operation of electrical devices and systems. The high current calculated in this example serves as a reminder of the potential hazards associated with electricity and the importance of proper safety measures and circuit protection.