What is the total capacitance of two capacitors in series, C1 = 4μF and C2 = 6μF

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Question: What is the total capacitance of two capacitors in series, C1 = 4μF and C2 = 6μF?

Options:

Correct Answer: 2.4μF

Solution:

The formula for total capacitance in series is 1/C_total = 1/C1 + 1/C2. Thus, 1/C_total = 1/4 + 1/6 = 5/12, so C_total = 12/5 = 2.4μF.

What is the total capacitance of two capacitors in series, C1 = 4μF and C2 = 6μF?

What is the total capacitance of two capacitors in series, C1 = 4μF and C2 = 6μF?

Step 1: Identify the values of the capacitors. Here, C1 = 4μF and C2 = 6μF.
Step 2: Write down the formula for total capacitance in series: 1/C_total = 1/C1 + 1/C2.
Step 3: Substitute the values of C1 and C2 into the formula: 1/C_total = 1/4 + 1/6.
Step 4: Find a common denominator to add the fractions. The common denominator for 4 and 6 is 12.
Step 5: Rewrite the fractions with the common denominator: 1/4 = 3/12 and 1/6 = 2/12.
Step 6: Add the fractions: 1/C_total = 3/12 + 2/12 = 5/12.
Step 7: To find C_total, take the reciprocal of 5/12: C_total = 12/5.
Step 8: Calculate 12/5 to get the final answer: C_total = 2.4μF.

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