NEET Combination of Capacitors MCQs & PYQs

The figure shows

$n$

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
   $C$ 

   .

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
     $n$ 

     capacitors in the last group.

3. Capacitance of a single group:
   • For
     $k$ 

     capacitors in series, the equivalent capacitance is: $$C_k = \frac{C}{k}$$ 

4. Net capacitance:
   • These groups are connected in parallel. The total equivalent capacitance
     $C_{eq}$ 

     is the sum of the capacitances of all groups: $$C_{eq} = \sum_{k=1}^{n} C_k = \sum_{k=1}^{n} \frac{C}{k}$$ 

5. Simplify:
   • The sum of the reciprocals of integers up to
     $n$ 

     is: $$C_{eq} = C \cdot \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \right)$$ 

   • After simplifications, the given result:
     $$C_{eq} = \frac{(n+1)n}{2}C$$ 

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

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

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

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.

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

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

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

So, Charge on  the capacitors in CE.

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
   $C$ 

   at the top and bottom in parallel.

2. For capacitors in parallel, the equivalent capacitance is simply the sum of their capacitances:
   $$C_{\text{eq}} = C + C = 2C$$ 

Final Answer:

The equivalent capacitance of the given combination is 2C.

For the common potential to be zero, the initial charges on the capacitors must be equal in magnitude and connected with opposite polarities: \( q_1 = q_2 \Rightarrow 120 C_1 = 200 C_2 \Rightarrow 3 C_1 = 5 C_2 \).

Since both capacitors are in series and have equal capacitance, the total potential difference divides equally between them. The maximum potential is limited by the weaker capacitor: \( V_{\text{max}} = 2 \times 4\text{V} = 8\text{V} \).

When widely separated conductors are connected, they are effectively in parallel as they share a common potential. The equivalent capacitance is the sum of individual capacitances: \(C_{eq} = C_1 + C_2 = 4 \pi \epsilon_0 R_1 + 4 \pi \epsilon_0 R_2 = 4 \pi \epsilon_0 (R_1 + R_2)\).

For each capacitor, calculate the maximum charge it can hold: \(Q_1 = C_1 V_1 = (2 \mu\text{F})(3\text{V}) = 6 \mu\text{C}\), \(Q_2 = C_2 V_2 = (3 \mu\text{F})(2\text{V}) = 6 \mu\text{C}\), \(Q_3 = C_3 V_3 = (5 \mu\text{F})(1text{V}) = 5 \mu\text{C}\). In a series combination, the charge must be the same on each capacitor, so the maximum charge is the minimum of these, \(Q_{max} = 5 \mu\text{C}\). The equivalent capacitance in series is \(\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} = \frac{15+10+6}{30} = \frac{31}{30} \text{F}^{-1}\), so \(C_{eq} = \frac{30}{31} \mu\text{F}\). The maximum voltage is \(V_{max} = \frac{Q_{max}}{C_{eq}} = \frac{5 \mu\text{C}}{(30/31) \mu\text{F}} = \frac{5 \times 31}{30} = \frac{31}{6} \text{ Volts}\).

Let \(C\) be the capacitance of each capacitor. For capacitor A, initial energy \(U_A = \frac{Q_A^2}{2C} = 4\text{J}\), so \(Q_A = \sqrt{8C}\). Capacitor B is uncharged, so \(Q_B = 0\). When connected in parallel, total charge is conserved: \(Q_{total} = Q_A + Q_B = \sqrt{8C}\). The equivalent capacitance is \(C_{eq} = C + C = 2C\). The final total energy is \(U_{total} = \frac{Q_{total}^2}{2C_{eq}} = \frac{(\sqrt{8C})^2}{2(2C)} = \frac{8C}{4C} = 2\text{J}\).