Seven capacitors each of capacitance 2$\mu$F are to be connected in a configuration to obtain an effective capacitance (10/11)$\mu$F. The combination is \dots
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When designing circuits, if the target equivalent capacitance ($10/11 \approx 0.9$) is significantly smaller than the individual capacitance ($2.0$), the dominant macro-structure of the circuit must be a series configuration.
Step 1: Understanding the Concept:
Capacitors in parallel add up ($C_p = C_1 + C_2 \dots$), while in series they follow the reciprocal rule ($1/C_s = 1/C_1 + 1/C_2 \dots$). Step 2: Formula Application:
To get a denominator like 11, we likely need a parallel combination of 5 capacitors ($5 \times 2 = 10\mu$F).
Let this $10\mu$F block be in series with the remaining 2 capacitors ($2\mu$F each). Step 3: Explanation:
$1/C_{eq} = 1/10 + 1/2 + 1/2 = 1/10 + 5/10 + 5/10 = 11/10$.
$C_{eq} = 10/11 \mu$F. Step 4: Final Answer:
The correct combination is 5 capacitors in parallel connected in series with 2 individual capacitors.
