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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

Arun
25763 Points
one year ago
First of all We need to take out Potential on one Capaciter i.e :

=> Potential on each capacitor V = 1000/n
Here 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 = 8
m = 8 ( 8 rows )

➡➡ Minimum number of capacitor = 8 × 4 =

♦ 32 capacitors . is the required

Vikas TU
14149 Points
one year ago
Dear student
Here is short cut.
Series grouping = 1/c1 + 1/c2
1/ceq = 1/8 + 1/8 + 1/8 + 1/8