Two parallel plate capacitors are identical, except that one of them is empty and the other contains a material with a dielectric constant of 4.2 in the space between the plates. The empty capacitor is connected between the terminals of an ac generator that has a fixed frequency and rms voltage. The generator delivers a current of 0.23 A. What current does the generator deliver after the other capacitor is connected in parallel with the first one?
Help :/
Vm = Xc Im
Xc = 1/ωC
therefore for fixed V
I2/I1 = C2 /C1
C2 = 4.2C1 + C1 = 5.2C1 (they are connected in parallel)
I2 = (C2/C1)I1 = 5.2*0.22 = 1.144 A
Oh, it's a shocking question! Let me try to lighten the mood with some humor.
Well, when the second capacitor is connected in parallel, it's like having "double the pleasure, double the fun" for the generator. Since the capacitors are identical, the second one with the dielectric constant of 4.2 will simply enhance the overall capacitance.
So, if the generator initially delivers a current of 0.23 A with just the empty capacitor, now with the second capacitor being added, it's like a tag-team of capacitors working together. This increased capacitance will have a direct impact on the current delivered by the generator.
But, before we calculate the new current, let's take a moment to savor the thought of these two capacitors, happily paired together, buzzing with excitement, ready to "amplify" their impact!
Alright, now back to the question. To determine the new current, we need to keep in mind that the current flowing through a capacitor is inversely proportional to the capacitance. So, when the capacitance increases, the current decreases.
In this case, since the dielectric constant of the material is 4.2, we can assume that the capacitance of the second capacitor is 4.2 times higher than the empty one. Therefore, the new current delivered by the generator would be 0.23 A divided by 4.2.
I hope this shocking answer helps you out! Remember, laughter is the best conductor of knowledge!
To solve this problem, we need to use the concept of capacitors connected in parallel and the relationship between capacitance and current in an AC circuit.
1. Capacitors connected in parallel have the same voltage across them. Since the empty capacitor is already connected to the AC generator, the voltage across it remains unchanged.
2. The capacitance of a capacitor with a dielectric material is given by the equation:
C' = k * C,
where C' is the capacitance with dielectric, k is the dielectric constant, and C is the capacitance without dielectric.
3. The current in an AC circuit containing a capacitor is given by the equation:
I = V * ω * C,
where I is the current, V is the voltage, ω is the angular frequency, and C is the capacitance.
4. Since the generator has a fixed frequency, ω remains constant. We can also assume that the resistance in the circuit remains constant.
Now, let's solve the problem step-by-step:
Step 1: Determine the current delivered by the generator with the empty capacitor.
Given:
Current with empty capacitor (I1) = 0.23 A
Step 2: Calculate the capacitance of the empty capacitor (C1).
Since the capacitors are identical, the capacitance of the empty capacitor is the same as the capacitance of the capacitor with the dielectric material.
Step 3: Determine the current delivered by the generator when the second capacitor is connected in parallel.
To find the current, we need to calculate the capacitance of the capacitor with the dielectric material (C2) and then use the formula:
I2 = V * ω * C2,
where I2 is the current when the second capacitor is connected.
Step 4: Calculate the capacitance of the capacitor with the dielectric material (C2).
The capacitance with the dielectric material is given by:
C2 = k * C1,
where k is the dielectric constant.
Step 5: Calculate the current delivered by the generator when the second capacitor is connected (I2).
Using the formula:
I2 = V * ω * C2,
plug in the values and calculate the current.
Now, let's perform the calculations:
Given:
Current with empty capacitor (I1) = 0.23 A
Dielectric constant (k) = 4.2
Step 1: Determine the current delivered by the generator with the empty capacitor.
I1 = 0.23 A
Step 2: Calculate the capacitance of the empty capacitor (C1).
Since the capacitors are identical, the capacitance of the empty capacitor is the same as the capacitance of the capacitor with the dielectric material.
Step 3: Determine the current delivered by the generator when the second capacitor is connected in parallel.
I2 = ?
Step 4: Calculate the capacitance of the capacitor with the dielectric material (C2).
C2 = k * C1
Step 5: Calculate the current delivered by the generator when the second capacitor is connected (I2).
I2 = V * ω * C2
Now, plug in the values and calculate the current.
Note: We need the numerical values of the voltage and angular frequency to calculate the current accurately. Please provide those values for further calculations.
To determine the current delivered by the generator after the second capacitor is connected in parallel with the first one, we must understand the behavior of capacitors in a parallel circuit.
In a parallel circuit, the voltage across both capacitors remains the same. This means that if the generator provides a fixed rms voltage, it will remain the same with or without the second capacitor.
However, the current in a parallel circuit is divided among the components based on their capacitive reactance. The capacitive reactance depends on the capacitance and the frequency of the AC generator.
The formula for capacitive reactance (Xc) is given by:
Xc = 1 / (2πfC)
where f is the frequency of the AC generator and C is the capacitance.
Since the empty capacitor and the capacitor with the dielectric constant of 4.2 are identical, their capacitance values will be the same.
When the generator is connected to the empty capacitor, it delivers a current of 0.23 A. Let's call this current I1.
Now, let's denote the current delivered by the generator after the second capacitor is connected as I2.
Since the voltage across the capacitors remains the same, we can say:
I1 * Xc1 = I2 * Xc2
Since the capacitors are identical, Xc1 and Xc2 will be the same.
Therefore, we have:
I1 = I2 * Xc2 / Xc1
Now, let's consider the relationship between the capacitance and the dielectric constant. The capacitance with a dielectric material (C2) can be expressed as:
C2 = C1 * k
where C1 is the capacitance without the dielectric and k is the dielectric constant.
Substituting this into the equation for Xc, we have:
Xc2 = 1 / (2πfC2) = 1 / (2πf(C1 * k))
Xc1 = 1 / (2πfC1)
Substituting the values of Xc2 and Xc1 into the equation for I1, we have:
I1 = I2 * (1 / (2πf(C1 * k))) / (1 / (2πfC1))
I1 = I2 * (C1 / (C1 * k))
Simplifying:
I2 = I1 * (C1 * k) / C1
Since C1 / C1 = 1, the equation becomes:
I2 = I1 * k
Therefore, the current delivered by the generator after the second capacitor is connected will be the current delivered initially (0.23 A) multiplied by the dielectric constant (4.2):
I2 = 0.23 A * 4.2
I2 = 0.966 A
Hence, the generator will deliver a current of approximately 0.966 A after the second capacitor is connected in parallel with the first one.