Answer :
The equivalent capacitance at terminals a and b is approximately 2.17 Farads (F).
How can we determine the equivalent capacitance in this circuit?
To determine the equivalent capacitance, we need to consider the arrangement of the capacitors. In this case, the capacitors are connected in parallel and series combinations. Let's break down the steps:
Capacitors C1, C2, C3, C4, and C5 are connected in parallel. When capacitors are connected in parallel, their capacitances add up. Therefore, the total capacitance of C1 to C5 is 12F + 4F + 17F + 9F + 15F = 57F.
Now, we have an equivalent capacitor formed by C1 to C5 with a capacitance of 57F. This equivalent capacitor, let's call it CEQ1, is in series with C3. When capacitors are connected in series, their reciprocals add up. So, the reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of CEQ1 and C3.
Using the formula for capacitors in series: 1/CEQ2 = 1/CEQ1 + 1/C3
Substituting the values: 1/CEQ2 = 1/57F + 1/17F
Solving for CEQ2: CEQ2 ≈ 7.22F
Finally, we have CEQ2 in series with C4 and C5. Again, we apply the formula for capacitors in series to find the equivalent capacitance.
1/CEQ3 = 1/CEQ2 + 1/C4 + 1/C5
Substituting the values: 1/CEQ3 = 1/7.22F + 1/9F + 1/15F
Solving for CEQ3: CEQ3 ≈ 2.17F
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