Calculating Charge And Voltage In A Resistor Circuit
Hey guys! Let's dive into a super interesting physics problem today. We're going to break down how to calculate the charge flowing through a resistor and the voltage across specific points in a circuit. It might sound intimidating, but trust me, we'll make it easy to understand. So, grab your thinking caps, and let's get started!
Understanding the Circuit and Key Concepts
To really nail these calculations, we first need to visualize the circuit and wrap our heads around some crucial concepts. Imagine a circuit with resistors – these are like little roadblocks that resist the flow of electrical current. In our case, we have resistors of 4Ω, 6Ω, and 12Ω. The symbol 'Ω' stands for ohms, which is the unit of resistance. Think of resistance like friction in a pipe; the higher the resistance, the harder it is for current to flow.
Now, imagine we apply a 12-volt potential difference (that's like electrical pressure) across points A and B in the circuit. This potential difference, measured in volts (V), pushes the current through the circuit. The higher the voltage, the more 'push' there is.
Current, measured in amperes (A), is the flow of electrical charge. It's like the amount of water flowing through a pipe. Charge, measured in coulombs (C), is the fundamental electrical property that causes the flow of current. Think of coulombs as the individual 'bits' of electricity that are moving.
Finally, we need to remember Ohm's Law, a fundamental principle that connects voltage (V), current (I), and resistance (R): V = IR. This simple equation is our superpower for solving circuit problems! It tells us that the voltage across a resistor is equal to the current flowing through it multiplied by its resistance. Armed with these concepts, we're ready to tackle the specific questions.
Breaking Down the First Question: Charge Flow in the 12Ω Resistor
Our first mission is to figure out the charge flowing through the 12Ω resistor in 1 minute. To do this, we'll need to use Ohm's Law and a bit of time-related magic.
First things first, let's find the current flowing through the 12Ω resistor. But wait, we don't have the voltage across it directly! This is where we need to analyze the circuit configuration. If the 12Ω resistor is in series with other resistors, the current flowing through it is the same as the current flowing through the entire series combination. If it's in parallel, we'll need to calculate the voltage across the parallel branch first.
Let's assume, for example, that the 6Ω and 12Ω resistors are in parallel. This means they share the same voltage. To find this voltage, we first need to find the equivalent resistance of the parallel combination. The formula for the equivalent resistance (Rp) of two resistors in parallel is:
1/Rp = 1/R1 + 1/R2
In our case, R1 = 6Ω and R2 = 12Ω. Plugging these values in, we get:
1/Rp = 1/6 + 1/12 = 3/12
So, Rp = 12/3 = 4Ω.
Now, if this 4Ω equivalent resistance is in series with the 4Ω resistor (the one mentioned in the problem), the total resistance (Rt) of the circuit is:
Rt = 4Ω + 4Ω = 8Ω
Now we can use Ohm's Law to find the total current (It) flowing through the circuit:
It = V / Rt = 12V / 8Ω = 1.5A
Since the 4Ω resistor and the parallel combination of 6Ω and 12Ω are in series, this 1.5A is the current flowing through the equivalent 4Ω resistance (Rp). To find the voltage (Vp) across the parallel combination (and thus across the 12Ω resistor), we use Ohm's Law again:
Vp = It * Rp = 1.5A * 4Ω = 6V
Great! Now we know the voltage across the 12Ω resistor is 6V. We can finally calculate the current (I12) flowing through it:
I12 = Vp / R = 6V / 12Ω = 0.5A
Phew! That was a bit of a journey, but we've got the current. Now, to find the charge (Q) flowing in 1 minute (which is 60 seconds), we use the relationship between current, charge, and time:
I = Q / t
Rearranging for Q, we get:
Q = I * t = 0.5A * 60s = 30C
So, the charge flowing through the 12Ω resistor in 1 minute is 30 coulombs! Awesome!
Tackling the Second Question: Voltage VBC
The second part of our mission is to find the voltage VBC. This is the potential difference between points B and C in our circuit. Remember, we've already assumed a configuration where the 6Ω and 12Ω resistors are in parallel. Point B would be at the junction before these resistors split, and point C would be at the other end of the 12Ω resistor.
We've already calculated the voltage across the parallel combination (and therefore across the 12Ω resistor) as Vp = 6V. This voltage, Vp, is precisely the voltage VBC we're looking for!
Therefore, VBC = 6V.
See? It wasn't so bad! By breaking down the problem into smaller steps and applying Ohm's Law, we were able to calculate both the charge flowing through the resistor and the voltage between two points in the circuit.
Key Formulas and Concepts Recap
Let's quickly recap the key formulas and concepts we used:
- Ohm's Law: V = IR (Voltage = Current * Resistance)
- Equivalent Resistance in Parallel: 1/Rp = 1/R1 + 1/R2
- Current, Charge, and Time: I = Q / t (Current = Charge / Time)
- Series vs. Parallel Circuits: Understanding how resistors are connected is crucial for calculating currents and voltages.
Why This Matters: Real-World Applications
Now, you might be thinking,
