Combination of Capacitors – Page 2 – Rankers Physics
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Combination of Capacitors

Question 1:

A number of capacitors, each of equal capacitance C, are arranged as shown in Fig. The equivalent capacitance between A and B is:

n²C
(2n + 1) C
\[\frac{\left( n-1 \right)n}{2}C\]
\[\frac{\left( n+1 \right)n}{2}C\]
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Solution:

The figure shows nn groups of capacitors arranged in a specific pattern. Here's the reasoning for the given answer:

Solution:

  1. Each group consists of a series arrangement of capacitors with equal capacitance CC.
  2. The number of capacitors in each successive group increases by one, forming a triangular pattern:
    • 1st group: 1 capacitor,
    • 2nd group: 2 capacitors in series,
    • 3rd group: 3 capacitors in series, and so on, up to nn capacitors in the last group.
  3. Capacitance of a single group:
    • For kk capacitors in series, the equivalent capacitance is: Ck=CkC_k = \frac{C}{k}
  4. Net capacitance:
    • These groups are connected in parallel. The total equivalent capacitance CeqC_{eq} is the sum of the capacitances of all groups: Ceq=∑k=1nCk=∑k=1nCkC_{eq} = \sum_{k=1}^{n} C_k = \sum_{k=1}^{n} \frac{C}{k}
  5. Simplify:
    • The sum of the reciprocals of integers up to nn is: Ceq=C⋅(1+12+13+⋯+1n)C_{eq} = C \cdot \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \right)
    • After simplifications, the given result: Ceq=(n+1)n2CC_{eq} = \frac{(n+1)n}{2}C

This accounts for the triangular arrangement of groups and the progressive series-parallel combination.

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Question 2:

Seven capacitors, each of capacitance 2 μF, are to be combined to obtain a capacitance of 10/11 μF. Which of the following combinations is possible?

2 in parallel, 5 in series
3 in parallel, 4 in series
4 in parallel, 3 in series
5 in parallel, 2 in series
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Solution:

We need to check each option separately. We 5 capacitors are connected  in parallel, 2 capacitors are connected in series.

Ceq= (5C×C/2)/ (5C+C/2)= 5C/11 = 5×2/11 = 10/11 μF.

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Question 3:

In the circuit shown, the effective capacitance between points X and Y is:

3.33 μF
1 μF
0.44 μF
none of these
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Solution:

In the Upper arm 6 μF and 3 μF capacitors are in series , so equivalent capacitance is 2μF. The 3μF capacitor ( circled one) can be removed as it is part of balanced wheat stone bridge.

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Question 4:

Calculate the charge on the second capacitor before and after switch in the circuit is closed :

CE/2, CE
0, 0
0, CE
CE, 0
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Solution:

When Switch is Open Equivalent Capacitance is Ceq=C/2

So, Charge on Both the capacitors in CE/2.

When Switch is Closed 1st Capacitor is short circuited .Now  Equivalent Capacitance is Ceq=C.

So, Charge on  the capacitors in CE.

 

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Question 5:

The equivalent capacitance of the combination shown in figure is :

C
2C
3/2
C/2
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Solution:

 

Reason for the Short Circuit:

The middle capacitor is bypassed by a conducting wire (short circuit). Hence, no voltage difference exists across the middle capacitor, and it can be ignored in the calculation.

Simplified Circuit:

  1. The circuit reduces to two capacitors CC at the top and bottom in parallel.
  2. For capacitors in parallel, the equivalent capacitance is simply the sum of their capacitances: Ceq=C+C=2CC_{\text{eq}} = C + C = 2C

Final Answer:

The equivalent capacitance of the given combination is 2C.

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