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Grade: 12
        
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
5 months ago

Answers : (2)

Arun
24489 Points
							
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
 
5 months ago
Vikas TU
11137 Points
							
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  
5 months ago
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