A) \[1\,\,\mu F,\]\[2\,\,\mu F\]
B) \[6\,\,\mu F,\]\[2\,\,\mu F\]
C) \[12\,\,\mu F,\]\[4\,\,\mu F\]
D) \[3\,\,\mu F,\]\[16\,\,\mu F\]
Correct Answer: C
Solution :
[c] \[{{C}_{s}}=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=3\] \[{{C}_{p}}={{C}_{1}}+{{C}_{2}}=16\]\[\therefore \]\[{{C}_{1}}{{C}_{2}}=48\] \[{{C}_{1}}-{{C}_{2}}=\sqrt{{{({{C}_{1}}+{{C}_{2}})}^{2}}-4\,{{C}_{1}}{{C}_{2}}}\] \[=\sqrt{{{16}^{2}}-4\times 48}=\sqrt{64}=8\] \[\therefore \] \[{{C}_{1}}=12\mu F\]and \[{{C}_{2}}=4\mu F\]
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