First slide

Capacitors and capacitance

Question

Seven capacitors each of capacity $2 μ F$ are to be so connected to have a total capacity $\frac{10}{11} μ F$ . Which will be the necessary figure as shown

Easy

A
B
C
D

Solution

Let say each capacitor has C capacitance.
There are five parallel capacitors and their equivalent capacitance is 5C
and other two are in series with this combination C/2.
Finally equivalent capacitance of 5C and C/2 which are in series is $C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = 5 C \times (C / 2) / (5 C + C / 2) = 5 C / 11$ ,
given $C = 2 μ F$ then $C_{e q} = \frac{5 (2)}{11} = \frac{10}{11} μ F$ .

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