Question

While working on a physics at school physics lab, you require a $4\mu F$ capacitor in a circuit, across a potential difference of $1kV$.Unfortunately,$4\mu F$ capacitor are out of stock in you lab, but $2\mu F$ capacitors which can withstand a potential difference of $400V$ are available in plenty. If you decide to use the $2\mu F$ capacitor in place of $4\mu F$ capacitor, minimum number of capacitor required are$?$
A.$16$
B.$18$
C.$20$
D.$12$

Answer 1

Hint: Here, we know the potential difference value and capacitor, now we can find the minimum number of capacitors needed. According to that the total capacitance of this single equivalent capacitor depends on both the individual capacitors. So, by using some capacitor formula, we measure the necessary capacitor.
Useful formula:
Potential difference,
$V = IR$
Where,
$I$ is current
$V$ is voltage
$R$ is resistance
Capacitor,
$q = CV$
$q$ is charge
$C$ capacitance

Complete answer:
Given by,
$4\mu F$ capacitor in a circuit across a Potential difference of $1kV$
$2\mu F$ which can withstand a Potential difference of $400V$ to have $PD \geqslant 1000$
we need $3$ capacitors in series,
Here,
$\left( {3 * 400 = 1200V = 1.2kV > 1kV} \right)$
Capacitance of $3$ capacitors of $2\mu F$,
In Series,
 we get $2/3\mu F$
Now, we have capacitance of $4\mu F$
So,
We need to add $3$ capacitors in parallel,
Here,
Combine the equation,
$4/\left( {2/3} \right) = 12/2 = 6$
Then, total capacitors required,
We get,
$3*6 = 18$
Hence,
The Minimum number of capacitors required are $18$

Thus, Option B is the correct answer.

Note:
You add up the individual capacitances to measure the total overall capacitance of several connected capacitors in this manner. Thus, a mixture of series and parallel capacitors is required to obtain. In parallel combination, the minimum that can be obtained is when two condensers are linked in parallel.