Ten capacitors each of 2 µF capacity were arranged in two rows such that first row contains 6 capacitors in series where as the rest were in second row. The equivalent capacitance of the network is:
- \(\frac{5}{6}μF\)
- 20μF
- \(\frac{24}{5}μF\)
- \(\frac{6}{5}μF\)
The Correct Option is A
Solution and Explanation
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