Answer:

Explanation:
Given that,
Capacitance 1, 
Capacitance 2, 
Capacitance 3, 
C₁ and C₂ are connected in series. Their equivalent is given by :



Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :



So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.
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