To form a composite $16\,\mu F,\;1000\,V$ capacitor from a supply of identical capacitors marked $8\,\mu F,\;250\,V$, we require a minimum number of capacitors
AIIMS 2000, Diffcult
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(b) Suppose $C = 8\, µF ,\, C' = 16\,µF$
and $V = 250\, V,\, V' = 1000\,V$
Suppose $m$ rows of given capacitors are connected in parallel and each row contains $n$ capacitors then potential difference across each capacitor $V = \frac{{V'}}{n}$ and equivalent capacitance of network $C' = \frac{{mC}}{n}$ on putting the values we get $n = 4$ and $m = 8$
Total capacitors $= n × m = 4 × 8 = 32$
Short Trick : For such type of problems number of capacitors = $\frac{{C'}}{C} \times {\left( {\frac{{V'}}{V}} \right)^2} = \frac{{16}}{8}{\left( {\frac{{1000}}{{250}}} \right)^2} = 32$
and $V = 250\, V,\, V' = 1000\,V$
Suppose $m$ rows of given capacitors are connected in parallel and each row contains $n$ capacitors then potential difference across each capacitor $V = \frac{{V'}}{n}$ and equivalent capacitance of network $C' = \frac{{mC}}{n}$ on putting the values we get $n = 4$ and $m = 8$
Total capacitors $= n × m = 4 × 8 = 32$
Short Trick : For such type of problems number of capacitors = $\frac{{C'}}{C} \times {\left( {\frac{{V'}}{V}} \right)^2} = \frac{{16}}{8}{\left( {\frac{{1000}}{{250}}} \right)^2} = 32$
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