Question

How many $6\;{\rm{\mu F}}$,$200\;{\rm{V}}$condensers are needed to make a condenser of $18\;{\rm{\mu F}}$, $600\;{\rm{V}}$?
A. 9
B. 18
C. 3
D. 27

Answer 1

The above problem can be resolved by taking the first three combinations of the given individual condensers connected in series. The obtained value of net capacitance needs to be connected in parallel for nine such capacitors to obtain the desired condenser of the capacitance 18 microfarad. The total condenser required is given by taking the capacitor's product required in series connection and the number of condensers with that in parallel combination.

Complete step by step answer:
The capacitance of individual condenser is, ${C_1} = 6\;{\rm{\mu F}}$.
The voltage supply to the condenser is, ${V_1} = 200\;{\rm{V}}$.
The capacitance of the condenser formed by the combination is, ${C_2} = 18\;{\rm{\mu F}}$.
The resultant voltage is, ${V_2} = 600\;{\rm{V}}$.
To obtain the net capacitance of $18\;{\rm{\mu F}}$, the combination of three1 individual capacitors need to be placed in the series combination as,

\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_1}}}\\\ \dfrac{1}{C} = \dfrac{3}{{6\;{\rm{\mu F}}}}\\\ C = 2\;{\rm{\mu F}} \end{array}$$ In order to get the net capacitance of $$18\;{\rm{\mu F}}$$, the given combination for the capacitance of $$2\;{\rm{\mu F}}$$should be repeated 9 times in parallel combination as, $$\begin{array}{l} {C_3} = 9 \times C\\\ {C_3} = 9 \times 2\;{\rm{\mu F}}\\\ {{\rm{C}}_3} = 18\;{\rm{\mu F}} \end{array}$$ Now, as the nine rows of capacitors are connected in parallel with the 3 capacitors in series. So, the total number of condensers required is, $$\begin{array}{l} n = 9 \times 3\\\ n = 27 \end{array}$$ Therefore, the total number of condensers required is 27 and option D is correct. **Note:** Try to understand the concept of capacitance along with the various combinations of capacitors like series and parallel combinations. In series combination, the net capacitance comes out to be lesser in magnitude, while in parallel combination, the result is increased numerically.