A capacitance of 2 mu F is required in an ele | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $n$ be the number of capacitors in each parallel row and $m$ be the number of such rows in series.Each capacitor can withstand $300\ V$. To wi

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(D) Let $n$ be the number of capacitors in each parallel row and $m$ be the number of such rows in series.Each capacitor can withstand $300\ V$. To withstand a total potential difference of $1000\ V$,the number of capacitors in series $(m)$ must satisfy $m 	imes 300 \ge 1000$,which gives $m \ge 3.33$. Thus,we need at least $m = 4$ rows in series.The potential difference across each row will be $1000/4 = 250\ V$,which is within the safe limit of $300\ V$.The equivalent capacitance of one row of $n$ capacitors in parallel is $C_{row} = n 	imes 1\ \mu F = n\ \mu F$.Since there are $m = 4$ such rows in series,the total equivalent capacitance is given by $\frac{1}{C_{eq}} = \frac{1}{C_{row}} + \frac{1}{C_{row}} + \frac{1}{C_{row}} + \frac{1}{C_{row}} = \frac{4}{n}$.Given $C_{eq} = 2\ \mu F$,we have $\frac{1}{2} = \frac{4}{n}$,which implies $n = 8$.Total number of capacitors = $m 	imes n = 4 	imes 8 = 32$.

$A$ capacitance of $2\ \mu F$ is required in an electrical circuit across a potential difference of $1.0\ kV$. $A$ large number of $1\ \mu F$ capacitors are available which can withstand a potential difference of not more than $300\ V$. The minimum number of capacitors required to achieve this is

Similar Questions

Capacitors of capacities $C_1$ and $C_2$ are connected in series. If the combination is connected to a supply of $V$ volt, then the potential difference across capacitor $C_2$ is:

The effective capacitances of two capacitors are $3\,\mu F$ and $16\,\mu F$,when they are connected in series and parallel respectively. The capacitances of the two capacitors are:

The value of equivalent capacitance of the combination shown in the figure between the points $P$ and $Q$ is

Three capacitors of $2.0\;\mu F$,$3.0\;\mu F$,and $6.0\;\mu F$ are connected in series to a $10\,V$ source. The charge on the $3.0\;\mu F$ capacitor is ........ $\mu C$.

When three capacitors of equal capacities are connected in parallel and one of the same capacity is connected in series with the combination,the resultant capacity is $4.5 \mu F$. The capacity of each capacitor is: (in $\mu F$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module