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a capacitance of 2 micro farad is required in an electrical circuit across a potential difference of 1.0kv. a large number of 1micro farad capacitors are available which can withstand a potential difference of not more than 300V.the minimum number of capacitors required to achieve this

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11 months ago

```							First of all We need to take out Potential on one Capaciter i.e : => Potential on each capacitor V = 1000/nHere V = 300 So ,, => 1000/n = 300 = n = 1000÷300 => n = 10/3 = 4 ( Approximately )=> C( eq ) = C/n × m Substituting the values of C(eq) and C we get ,, (( n capacitors ,, m rows )) => 1/n × m = 2 Therefore ,, m = n × 2 = 4 × 2 = 8m = 8 ( 8 rows ) ➡➡ Minimum number of capacitor = 8 × 4 = ♦ 32 capacitors . is the required
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11 months ago
```							Dear student Here is short cut.Series grouping = 1/c1 + 1/c2 1/ceq = 1/8 + 1/8 + 1/8 + 1/8 Ceq = 2micro farad Total capacitance required = 8*4 = 32 Good Luck
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11 months ago
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