Capacitors: Practice Problem & Solution
Three capacitors each of capacity $4 \mu F$ are to be connected in such a way that the effective capacitance is $6 \mu F$. This can be done by: (2003)
Solution Explained:
To solve this problem, we apply the core principles of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:
Connecting two $4 \mu F$ capacitors in series gives an equivalent capacitance of $2 \mu F$. Adding the third $4 \mu F$ capacitor in parallel gives a total of $2 \mu F + 4 \mu F = 6 \mu F$.
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