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Grade 12Electric Current

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

GAYATRI RAMA SRIJA GUNDABATTULA

6 Years agoGrade 12

2 Answers

Arun

6 Years 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

6 Years 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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