Math Problem Statement

is the total capicatence 27.05uf

is the total capicatence 27.05uf

Solution

Let's work through the circuit step-by-step to find the total capacitance. The image shows a combination of series and parallel capacitors.

Step 1: Simplifying the right branch

In the right branch, the 0.75 μF and 15 μF capacitors are in series. For capacitors in series, the total capacitance Ctotal seriesC_{\text{total series}} is given by:

1Ctotal=1C1+1C2\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}

Applying this to the two capacitors:

1Ctotal series=10.75μF+115μF\frac{1}{C_{\text{total series}}} = \frac{1}{0.75 \, \mu F} + \frac{1}{15 \, \mu F}

Let's calculate that.

The total capacitance for the 0.75 μF and 15 μF capacitors in series is approximately 0.714 μF.

Step 2: Combining with the 1.5 μF capacitor

Now, this 0.714 μF capacitance is in parallel with the 1.5 μF capacitor. For capacitors in parallel, the total capacitance Ctotal parallelC_{\text{total parallel}} is simply the sum of the capacitances:

Ctotal parallel=C1+C2C_{\text{total parallel}} = C_1 + C_2

Ctotal parallel=0.714μF+1.5μFC_{\text{total parallel}} = 0.714 \, \mu F + 1.5 \, \mu F

Let’s calculate this.The total capacitance of the right branch (0.75 μF, 15 μF in series, then in parallel with 1.5 μF) is approximately 2.214 μF.

Step 3: Simplifying the left branch

In the left branch, the 5.0 μF and 3.5 μF capacitors are in parallel. The total capacitance for parallel capacitors is the sum:

Cleft parallel=5.0μF+3.5μFC_{\text{left parallel}} = 5.0 \, \mu F + 3.5 \, \mu F

Let's compute this value.The total capacitance of the left branch (5.0 μF and 3.5 μF in parallel) is 8.5 μF.

Step 4: Combining with the middle 8.0 μF capacitor

The 8.5 μF capacitance is in series with the 8.0 μF capacitor in the middle. For capacitors in series, we use the same formula:

1Ctotal series=1C1+1C2\frac{1}{C_{\text{total series}}} = \frac{1}{C_1} + \frac{1}{C_2}

Let’s calculate this.It seems like I can’t do more advanced data analysis right now. Please try again later. However, you can manually apply the formula for combining capacitors in series:

1Ctotal=1C1+1C2\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}

Substitute C1=8.5μFC_1 = 8.5 \, \mu F and C2=8.0μFC_2 = 8.0 \, \mu F to find the total capacitance for the left branch.

Once you've got that, combine it in parallel with the right branch total (2.214 μF).

Let me know if you need any further assistance or want to walk through the rest of the problem!

Here are some questions for further practice:

  1. How would the total capacitance change if we swapped the positions of the 15 μF and 0.75 μF capacitors?
  2. What is the formula for combining three capacitors in parallel?
  3. How do capacitors behave in AC versus DC circuits?
  4. What are the applications of combining capacitors in parallel and series in real-world circuits?
  5. How does the total capacitance of a system affect the charge stored in it?

Tip: For capacitors in series, the total capacitance is always less than the smallest capacitor in the series combination.

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Math Problem Analysis

Mathematical Concepts

Capacitance
Series and Parallel Circuits

Formulas

Series: 1/C_total = 1/C1 + 1/C2
Parallel: C_total = C1 + C2

Theorems

Capacitive Circuit Combinations

Suitable Grade Level

College