If two capacitors of 4 µF and 6 µF are connected in series, what is the total ca

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Question: If two capacitors of 4 µF and 6 µF are connected in series, what is the total capacitance?

Options:

Correct Answer: 2.4 µF

Solution:

1/C_total = 1/C1 + 1/C2 = 1/4 + 1/6 = 5/12, thus C_total = 12/5 = 2.4 µF.

If two capacitors of 4 µF and 6 µF are connected in series, what is the total capacitance?

If two capacitors of 4 µF and 6 µF are connected in series, what is the total capacitance?

Step 1: Identify the values of the capacitors. We have C1 = 4 µF and C2 = 6 µF.
Step 2: Use the formula for total capacitance in series: 1/C_total = 1/C1 + 1/C2.
Step 3: Substitute the values 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: 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: C_total = 12/5.
Step 8: Calculate 12/5 to get the final answer: C_total = 2.4 µF.

Capacitance in Series – When capacitors are connected in series, the total capacitance is found using the formula 1/C_total = 1/C1 + 1/C2.