Circuit Analysis: Calculating Current In Resistor

by Elias Adebayo 50 views

Hey guys! Today, we're diving into a classic physics problem involving batteries and resistors. We're going to break down a circuit, understand how the components interact, and ultimately calculate the current flowing through a specific resistor. So, buckle up and let's get started!

The Circuit Scenario

Imagine a circuit with two batteries connected in a rather interesting way. We have a powerful 24V battery with an internal resistance of 2Ω. This battery is our main power source. Then, we have a smaller 6V battery, also with an internal resistance of 2Ω. These batteries are connected in a way that their voltages oppose each other. Finally, we have an external resistor of 4Ω in the mix. Our mission, should we choose to accept it, is to determine the current flowing through this 4Ω resistor.

Understanding the Components

Before we jump into calculations, let's make sure we're all on the same page about the components in our circuit:

  • Batteries: Batteries are the heart of our circuit, providing the electromotive force (EMF) that drives the current. Think of EMF as the electrical 'push' that gets the charges moving. But, real-world batteries aren't perfect. They have internal resistance, which acts like a tiny resistor inside the battery itself, hindering the current flow. Our 24V battery has a larger EMF but the same internal resistance as the 6V battery.
  • Resistors: Resistors are the brakes in our circuit, opposing the flow of current. The 4Ω resistor is our target – we want to know how much current is flowing through it.

Setting Up the Problem

Now, let's visualize the circuit. We have the 24V battery pushing current in one direction, and the 6V battery pushing current in the opposite direction. This opposition is crucial to understanding the overall current flow. The internal resistances of the batteries and the external 4Ω resistor all contribute to limiting the current.

To solve this, we'll use Kirchhoff's laws, which are fundamental principles for analyzing circuits. Kirchhoff's laws give us a systematic way to handle complex circuits with multiple loops and branches.

Applying Kirchhoff's Laws

Kirchhoff's laws come in two flavors:

  1. Kirchhoff's Current Law (KCL): This law states that the total current entering a junction (a point where wires connect) must equal the total current leaving the junction. It's like saying that the water flowing into a pipe system must equal the water flowing out.
  2. Kirchhoff's Voltage Law (KVL): This law states that the sum of the voltage drops around any closed loop in a circuit must equal zero. This is a consequence of the conservation of energy – the energy gained from the batteries must equal the energy lost across the resistors.

For our circuit, we'll primarily use KVL. We'll trace a loop around the circuit, accounting for the voltage gains from the batteries and the voltage drops across the resistors. Let's assume the current flowing through the circuit is 'I'.

Tracing the Loop

Starting from a point in the circuit, let's trace a loop in the direction of the assumed current flow. We encounter the following:

  • 24V Battery: We gain 24V as we move through the battery in the direction of its EMF.
  • 2Ω Internal Resistance (of the 24V battery): We drop a voltage of 2I across this resistor (using Ohm's Law: V = IR).
  • 4Ω Resistor: We drop a voltage of 4I across this resistor.
  • 2Ω Internal Resistance (of the 6V battery): We drop a voltage of 2I across this resistor.
  • 6V Battery: We lose 6V as we move through the battery against the direction of its EMF.

Setting Up the Equation

Now, we can apply KVL: The sum of the voltage gains and drops around the loop must equal zero.

So, we have:

24V - 2I - 4I - 2I - 6V = 0

Solving for the Current

Let's simplify the equation:

18V - 8I = 0

Now, solve for I:

8I = 18V

I = 18V / 8Ω

I = 2.25A

Therefore, the current flowing through the circuit is 2.25 Amperes.

Finding the Current Through the 4Ω Resistor

Since the components are connected in series, the current flowing through the 4Ω resistor is the same as the total current flowing through the circuit. So, the current through the 4Ω resistor is also 2.25A.

Conclusion

We've successfully analyzed the circuit and calculated the current flowing through the 4Ω resistor. We used Kirchhoff's laws, Ohm's Law, and a bit of circuit analysis to arrive at our solution. Remember, understanding the behavior of circuits is crucial in many areas of physics and engineering. I hope you guys found this helpful and insightful!

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Final Answer: The Final Answer is 2.25A\boxed{2.25 A}