An electrician requires a capacitance of 6 mu | vedclass

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(D) To withstand a potential difference of $1.5 kV$ $(1500 V)$ using capacitors that can withstand $500 V$ each,the number of capacitors $(m)$ in each

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(D) To withstand a potential difference of $1.5 kV$ $(1500 V)$ using capacitors that can withstand $500 V$ each,the number of capacitors $(m)$ in each series row is:$m = \frac{1500 V}{500 V} = 3$The equivalent capacitance of one such row of $3$ capacitors (each $2 \mu F$) is:$C_{row} = \frac{2 \mu F}{3}$If we connect $n$ such rows in parallel to achieve a total capacitance of $6 \mu F$,we have:$C_{eff} = n 	imes C_{row} = n 	imes \frac{2}{3} \mu F = 6 \mu F$$n = \frac{6 	imes 3}{2} = 9$The total number of capacitors required is $N = m 	imes n = 3 	imes 9 = 27$.

An electrician requires a capacitance of $6 \mu F$ in a circuit across a potential difference of $1.5 kV$. $A$ large number of $2 \mu F$ capacitors which can withstand a potential difference of not more than $500 V$ are available. The minimum number of capacitors required for the purpose is

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