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If two identical capacitors of 4 µF each are connected in series, what is the eq
If two identical capacitors of 4 µF each are connected in series, what is the equivalent capacitance?
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Practice Questions
1 question
Q1
If two identical capacitors of 4 µF each are connected in series, what is the equivalent capacitance?
2 µF
4 µF
6 µF
8 µF
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C_eq = 1 / (1/C1 + 1/C2) = 1 / (1/4 + 1/4) = 2 µF.
Questions & Step-by-step Solutions
1 item
Q
Q: If two identical capacitors of 4 µF each are connected in series, what is the equivalent capacitance?
Solution:
C_eq = 1 / (1/C1 + 1/C2) = 1 / (1/4 + 1/4) = 2 µF.
Steps: 6
Show Steps
Step 1: Identify the values of the capacitors. Here, both capacitors (C1 and C2) are 4 µF.
Step 2: Use the formula for equivalent capacitance in series: C_eq = 1 / (1/C1 + 1/C2).
Step 3: Substitute the values into the formula: C_eq = 1 / (1/4 + 1/4).
Step 4: Calculate the fractions: 1/4 + 1/4 = 2/4.
Step 5: Now, substitute back into the formula: C_eq = 1 / (2/4).
Step 6: Simplify the equation: C_eq = 1 / (0.5) = 2 µF.
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