We need to connect a capacitor of 16 mu F, 10 | vedclass

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

Question

(C) Given: Required capacitance $C' = 16\ \mu F$,Required voltage $V' = 1000\ V$. Available capacitance $C = 8\ \mu F$,Available voltage $V = 250\ V$.

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) Given: Required capacitance $C' = 16\ \mu F$,Required voltage $V' = 1000\ V$. Available capacitance $C = 8\ \mu F$,Available voltage $V = 250\ V$.To achieve the required voltage $V'$,we need to connect $n$ capacitors in series: $n = V' / V = 1000 / 250 = 4$.Now,the equivalent capacitance of $n$ capacitors in series is $C_{eq} = C / n = 8 / 4 = 2\ \mu F$.To achieve the required capacitance $C'$,we need to connect $m$ such series branches in parallel: $m = C' / C_{eq} = 16 / 2 = 8$.Total number of capacitors required = $m 	imes n = 8 	imes 4 = 32$.

We need to connect a capacitor of $16\ \mu F, 1000\ V$. However,we only have capacitors of $8\ \mu F, 250\ V$ available. How many such capacitors are required?

Similar Questions

Three capacitors each of $4\,\mu F$ are to be connected in such a way that the effective capacitance is $6\,\mu F$. This can be done by connecting them:

In the following circuit,the equivalent capacitance between terminal $A$ and terminal $B$ is: (in $\mu F$)

Obtain the equivalent capacitance of the network in the figure. For a $300 \; V$ supply,determine the charge and voltage across each capacitor.

The equivalent capacitance between points $A$ and $B$ in the figure shown below is .......... $\mu F$.

Two capacitors of capacitance $4\,\mu F$ and $6\,\mu F$ are connected in series. $A$ potential difference of $500\;V$ is applied to the outer plates of the two-capacitor system. The charge on each capacitor is numerically:

Vedclass Test Series

Exam Paper Generator

Online Exam Module