Question

To form a composite $16\mu F$, 1000 V capacitor from a supply of identical capacitors marked $8\mu F$, 250 V, we require a minimum number of n capacitors where n is:
A. 40
B. 32
C. 8
D. 2

Answer 1

Hint: In a series connection, charge is same and in a parallel connection, potential difference is same. So, we’ll find a minimum number of capacitors in series such that total potential difference is equal to the sum of individual potential differences and total capacitance is equal to the sum of individual capacitance in parallel $16\mu F$.

Complete step by step answer:
Let m be the number of capacitors connected in series and n be the number of capacitors connected in parallel. In a series connection, the value of charge remains same i.e., q is the same and the potential difference is the sum of individual potentials i.e$ = {V_1} + {V_2} + {V_3} + ............$$ + Vm$. So, in this case the individual potential of each capacitor is 250 V because all the capacitors are identical and the total potential difference is 1000. Thus, we need four capacitors in series as $4 \times 250V = 1000V$. The resultant capacitance due to these four capacitors i.e., $\dfrac{q}{{Cs}} = \dfrac{q}{c} + \dfrac{q}{c} + \dfrac{q}{c} + \dfrac{q}{c}$ Hence, $\dfrac{1}{{Cs}} = \dfrac{4}{8}$ i.e., $Cs = 2\mu F$.
In a parallel connection, the value of potential difference at their ends remains the same and the value of charge (q) changes i.e., total charge (Q) = sum of the charges due to individual charge. $Q = {q_1} + {q_2} + {q_3} + ........$ $ + {q_n}$which equals to $CV = cV + cV + .........$ $ + cV$ because all the capacitors are identical. $CV = ncV$ i.e., $C = nc$. The total capacitance required is $16\mu F$ and the value of each capacitance is $2\mu F$. Hence, the no. of capacitance required is 8 that are to be connected in parallel. Thus, four capacitors in series in each row are to be connected in 8 parallel columns to get and 1000V.
Thus $m = 4$ capacitance are connected in series in row and $n = 8$ are connected in parallel. Therefore, minimum numbers of capacitances are $m \times n = 4 \times 8 = 32$ capacitance.

So, the correct answer is “Option B”.

Note:
When the capacitances are connected in series, charge remains same and potential differences are equally divided to each of them if they are identical and when the capacitance is connected in parallel, potential difference is same across the ends and charge varies.