A) \[3\,\mu F\]
B) \[6\,\mu F\]
C) \[9\,\mu F\]
D) \[12\,\mu F\]
Correct Answer: A
Solution :
The last three capacitors oh the right, each of capacitance \[\operatorname{C}=9\,\mu F\] are in series and are equivalent to a capacitance C? given by \[\frac{1}{C'}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{1}{3}\] Or \[C'=3\,\mu F\] Since C? is in parallel with\[{{C}_{1}}\], the equivalent capacitance of the last part of the network is \[\operatorname{C}''=C'+{{C}_{1}}=3+6=9\,\mu F\] Continuing the process of calculation towards the left we notice that we are finally left with the combination whose equivalent capacitance is\[3\,\mu F\].
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