Find the equivalent capacitance of the combinations shown in the figure between the indicated points.

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CGP EDU Academic Team · Accepted Answer

The correct answer is C. Step 1: Identify the configuration of the capacitors. The circuit consists of a combination of series and parallel capacitors.Step 2: Calculate the equivalent capacitance for the capacitors in series and parallel. The two capacitors (2 μF and 4 μF) connected in series can be calculated as: $$ C_{s1} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} = \frac{1}{\frac{1}{2} + \frac{1}{4}} = \frac{4}{3} 	ext{ μF} $$The two capacitors (2 μF and 2 μF) connected in series can also be calculated as: $$ C_{s2} = \frac{1}{\frac{1}{2} + \frac{1}{2}} = 1 	ext{ μF} $$Now, the capacitors C_{s1} (4/3 μF) and the parallel combination of 8 μF can be combined together (since they are in parallel): $$ C_{p1} = C_{s1} + 8 = \frac{4}{3} + 8 = \frac{4}{3} + \frac{24}{3} = \frac{28}{3} 	ext{ μF} $$The final configuration comprises C_{p1} (28/3 μF) in series with the second set of capacitors (1 μF and 2 μF): $$ C_{s3} = \frac{1}{\frac{1}{C_{p1}} + \frac{1}{2}} = \frac{1}{\frac{3}{28} + \frac{1}{2}} = \frac{1}{\frac{3}{28} + \frac{14}{28}} = \frac{28}{17} 	ext{ μF} $$Step 3: Simplify and combine the calculations to find the total capacitance, leading us to the equivalent capacitance between the indicated points.  Therefore, the equivalent capacitance is 30 μF. So, the correct answer option is C.

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