A) connecting two in series and one in parallel
B) connecting two in parallel and one in series
C) connecting all of diem in series
D) connecting all of them in parallel
Correct Answer: A
Solution :
Key Idea: In series order, the net capacitance is, \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+......\] In parallel order, the net capacitance is, \[C={{C}_{1}}+{{C}_{2}}+{{C}_{3}}+.....\] We have given, \[{{C}_{1}}={{C}_{2}}={{C}_{3}}=4\mu F\] The network of three capacitors can be shown as Here, \[{{C}_{1}}\] and \[{{C}_{2}}\] are in series and the combination of two is in parallel with \[{{C}_{3}}\]. \[{{C}_{net}}=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}+{{C}_{3}}\] \[=\left( \frac{4\times 4}{4+4} \right)+4\] \[=2+4=6\,\mu F\] The corresponding network is shown Here, \[{{C}_{1}}\] and \[{{C}_{2}}\] are in parallel and this combination is in series with \[{{C}_{3}}\]. So, \[{{C}_{net}}=\frac{({{C}_{1}}+{{C}_{2}})\times {{C}_{3}}}{({{C}_{1}}+{{C}_{2}})+{{C}_{3}}}\] \[=\frac{(4+4)\times 4}{(4+4)+4}=\frac{32}{12}=\frac{8}{3}\mu F\] The corresponding network is shown All the three are in series. \[So,\,\,\frac{1}{{{C}_{net}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\] \[=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\] \[\therefore C=\frac{4}{3}\mu F\] The corresponding network is shown All of them are in parallel. So, \[{{C}_{net}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] = 4 + 4 + 4 = 12 \[\mu \]F Hence, only choice (a) is correct.You need to login to perform this action.
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