You are given thirty two capacitors each having capacity $$4$$ μF. How do you connect all of them to prepare a composite capacity having capacitance $$8$$ μF?

Question

You are given thirty two capacitors each having capacity $$4$$ μF. How do you connect all of them to prepare a composite capacity having capacitance $$8$$ μF?

Infinity Learn · Accepted Answer

Let m condensers are joined in series and n such groups are in parallel.Capacity of m condensers connected in $$series1C$$′$$=1C1+1C2+$$…$$=mC$$ or C′$$=Cm=4mWhen$$ n such groups are connected in parallel, $$thenCeff=nC$$′$$=4nmGiven$$ that $$Geff=8$$, ∴$$8=4nmOr$$     $$nm=2$$  or  $$n=2m$$                         ...$$(1)Further$$                 $$mn=32$$                           ....$$(2)Now$$    m×$$2m=32$$  or  $$m=4and$$     n $$=$$ $$2$$ x $$4$$ $$=$$ $$8$$

You are given thirty two capacitors each having capacity 4 $μ$ F. How do you connect all of them to prepare a composite capacity having capacitance 8 $μ$ F?

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You are given thirty two capacitors each having capacity 4 μF. How do you connect all of them to prepare a composite capacity having capacitance 8 μF?

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You are given thirty two capacitors each having capacity 4 $μ$ F. How do you connect all of them to prepare a composite capacity having capacitance 8 $μ$ F?