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Question: What is the equivalent capacitance of the system of capacitors between A & B? ![](https://www.veda...

What is the equivalent capacitance of the system of capacitors between A & B?

(A) 72C\dfrac{7}{2}C
(B) 1.6C1.6C
(C) CC
(D) None

Explanation

Solution

To solve this question, we need to imagine a battery across the points where the equivalent capacitance is required. Then, identifying the series and parallel combinations, apply the suitable formulae for the equivalent capacitance.

Formula Used: The formulae used in this solution are
1Cs=1C1+1C2+...........\Rightarrow \dfrac{1}{{{C_s}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} +...........
Cp=C1+C2+...........\Rightarrow {C_p} = {C_1} + {C_2} +...........
Where Cs{C_s} is the equivalent series capacitance, and Cp{C_p} is the equivalent parallel capacitance of the capacitances C1{C_1}, C2{C_2}.

Complete step by step answer
In the given circuit diagram, the rightmost two capacitances are connected in series combination. So there equivalent capacitance is
1C1=1C+1C\Rightarrow \dfrac{1}{{{C_1}}} = \dfrac{1}{C} + \dfrac{1}{C}
1C1=2C\Rightarrow \dfrac{1}{{{C_1}}} = \dfrac{2}{C}
Taking the reciprocal, we get
C1=C2\Rightarrow {C_1} = \dfrac{C}{2}
So the given circuit diagram can be redrawn as

Now, as we can see in the above figure, the rightmost two capacitances are arranged in parallel combination. So their equivalent capacitance becomes
C2=C+C2\Rightarrow {C_2} = C + \dfrac{C}{2}
C2=3C2\Rightarrow {C_2} = \dfrac{{3C}}{2}
So the above circuit reduces to

Now, the rightmost two capacitors of capacitances CC and 3C2\dfrac{{3C}}{2} are in series combination. So their equivalent capacitance becomes
1C3=1C+23C\Rightarrow \dfrac{1}{{{C_3}}} = \dfrac{1}{C} + \dfrac{2}{{3C}}
1C3=53C\Rightarrow \dfrac{1}{{{C_3}}} = \dfrac{5}{{3C}}
Taking the reciprocal, we get
C3=3C5\Rightarrow {C_3} = \dfrac{{3C}}{5}
So the above circuit reduces to

Finally, as we can observe in the above circuit diagram, the two capacitances are arranged in parallel combination. So the equivalent capacitance across A and B becomes
C4=C+3C5\Rightarrow {C_4} = C + \dfrac{{3C}}{5}
C4=8C5=1.6C\Rightarrow {C_4} = \dfrac{{8C}}{5} = 1.6C
Thus the equivalent capacitance between A and B is equal to 1.6C1.6C.

Hence, the correct answer is option B.

Note
We should not get confused between the formula of the series and equivalent combinations of resistance and that of the capacitances. The reciprocal relation is valid for the series combination for capacitance and for the parallel combination of the resistances. And the direct addition is valid for the parallel combination for capacitance and for the series combination of the resistances.