Question

When a parallel combination of 2 capacitors of 100 pF each is connected across a series combination of 2 capacitors of 200 pF each, the effective capacitance is

• A. 600 pF
• B. \(\frac{400}{9}\) pF
• C. 200 pF
• D. 100 pF
• E. 300 pF

Hint:

For series capacitors of equal value, the result is the value divided by the number of capacitors (\(200/2 = 100\)). For parallel, just multiply by the number (\(100 \times 2 = 200\)).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The problem asks for the total capacitance when two distinct combinations are connected "across" each other, which implies the two resulting equivalent capacitors are in parallel.

Step 2: Key Formula or Approach:
1. Capacitors in Parallel: \(C_p = C_1 + C_2\).
2. Capacitors in Series: \(\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}\).

Step 3: Detailed Explanation:
1. First combination: Parallel of two 100 pF capacitors.
\[ C_{eq1} = 100 + 100 = 200 \text{ pF} \]
2. Second combination: Series of two 200 pF capacitors.
\[ C_{eq2} = \frac{200 \cdot 200}{200 + 200} = \frac{40000}{400} = 100 \text{ pF} \]
3. Final combination: The two equivalents are connected in parallel.
\[ C_{total} = C_{eq1} + C_{eq2} = 200 + 100 = 300 \text{ pF} \]

Step 4: Final Answer:
The effective capacitance is 300 pF.