Question

Find equivalent capacitance between points A and B:
[Assume each conducting plate is having same dimensions and neglect the thickness of the plate, $\dfrac {{\epsilon }_{0}A}{d} = 7\mu F$, where A is area of plates]

$A.7\mu F$
$B. 11\mu F$
$C. 12\mu F$
$D. 15\mu F$

Answer 1

Draw equivalent circuit of the given circuit. Then calculate equivalent capacitance for the combination. And then using that effective combination find equivalent capacitance between points A and B.

Formula used:
${C}_{eq}= C +\dfrac {C\left( \dfrac {C}{2} \right) }{C+ \dfrac {C}{2} }$
${C}_{AB}= C+ \dfrac {{C}_{eq}C}{{C}_{eq}+C}$

Complete step-by-step answer:
Given: $C=\dfrac {{\epsilon }_{0}A}{d} =7\mu F$
Equivalent circuit of the given circuit can be drawn as:

Equivalent capacitance for the combination can be calculated by,
${C}_{eq}=C + \dfrac {C\left( \dfrac {C}{2} \right) }{C + \dfrac {C}{2} }$
$\therefore {C}_{eq}= C + \dfrac {C}{3}$
$\therefore {C}_{eq}= \dfrac {4}{3} C$
Now by substituting value of C we get,
${C}_{eq}=\dfrac {28}{3}$

Now, equivalent capacitance between points A and B can be given by,
${C}_{AB}= C+ \dfrac {{C}_{eq}C}{{C}_{eq}+ C}$

Now, substituting the values in above equation we get,
${C}_{AB}= 7 +\dfrac {\dfrac {28}{3} \times 7}{\dfrac {28}{3} +7}$
$\therefore {C}_{AB}=7+ 4$
$\therefore {C}_{AB}= 11\mu F$
Therefore, the equivalent capacitance between points A and B is 11 $\mu F$ .

So, the correct answer is “Option B”.

Note: Draw equivalent circuit for every circuit you get. The equivalent circuit makes calculation much easier. Remember the formula to find equivalent connections in series and parallel are different. Such as equivalent series combination for capacitors and resistors is $\dfrac {1}{{C}_{eq}} =\dfrac {1}{{C}_{1}} +\dfrac {1}{{C}_{2}} +...+\dfrac {1}{{C}_{N}}$ and
${R}_{eq}={R}_{1}+{R}_{2}+...+{R}_{N}$ respectively. Vice-versa is the case for equivalent parallel connections.