Three capacitors of equal capacitance when
connected in series have a net capacitance of G.
When connected in parallel they have a capacitance of C2.
What is the value of C1/C2?

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Solution

For series combination,
$\frac{1}{C_{1}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} [Since all capacitors have equal capacitances] = \frac{3}{C} Or C_{1} = \frac{C}{3} - - - (i) For parallel combination, C_{2} = C + C + C = 3 C - - - (ii) Now, \frac{C_{1}}{C_{2}} = \frac{C / 3}{3 C} = \frac{C}{3} \times \frac{1}{3 C} = \frac{1}{9}$

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Three capacitors of equal capacitance whenconnected in series have a net capacitance of G.When connected in parallel they have a capacitance of C2.What is the value of C1/C2?

For series combination, 1C1=1C+1C+1C [ Since all capacitors have equal capacitances]=3COr C1=C3---(i)For parallel combination,C2=C+C+C=3C---(ii)Now,C1C2=C/33C=C3×13C=19

Three capacitors of equal capacitance when
connected in series have a net capacitance of G.
When connected in parallel they have a capacitance of C2.
What is the value of C1/C2?