Three capacitors each of capacitance 18\(\mu F\) are connected in series. Calculate the effective capacitance of the capacitors
Question 1 Report
Three capacitors each of capacitance 18\(\mu F\) are connected in series. Calculate the effective capacitance of the capacitors
Answer Details
When capacitors are connected in series, the effective capacitance of the combination is given by:
\[\frac{1}{C_{\text{eff}}}=\frac{1}{C_1}+\frac{1}{C_2}+...\frac{1}{C_n}\]
where \(C_1\), \(C_2\)...\(C_n\) are the individual capacitances of the capacitors connected in series.
In this case, we have three capacitors, each with capacitance 18\(\mu F\), connected in series. So we have:
\[\frac{1}{C_{\text{eff}}}=\frac{1}{18\mu F}+\frac{1}{18\mu F}+\frac{1}{18\mu F}=\frac{3}{18\mu F}\]
Simplifying the right-hand side, we get:
\[\frac{1}{C_{\text{eff}}}=\frac{1}{6\mu F}\]
Taking the reciprocal of both sides, we get:
\[C_{\text{eff}}=6\mu F\]
Therefore, the effective capacitance of the three capacitors in series is 6\(\mu F\).
Hence, the answer is: 6.00\(\mu F\)
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