NEET Combination of Capacitors MCQs & PYQs

First Branch has a capacitance of 1F , for second branch it is 1/2F , for third branch it is 1/4 F and so, on . As all these branches are in parallel

\[ C_{eq}=C_{1}+C_{2}+C_{3}+....\]

\[ C_{eq}= 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8}+....=\frac{1}{1-\frac{1}{2}}=2\mu F\]

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.

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: \(C_{eq} = C_1 + C_2 = 4 \pi \epsilon_0 R_1 + 4 \pi \epsilon_0 R_2 = 4 \pi \epsilon_0 (R_1 + R_2)\).

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 \(U_{total} = \frac{Q_{total}^2}{2C_{eq}} = \frac{(\sqrt{8C})^2}{2(2C)} = \frac{8C}{4C} = 2\text{J}\).

Initial energy \(U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2\). Common potential after connection \(V = \frac{CV_1 + CV_2}{C + C} = \frac{V_1 + V_2}{2}\) for like-polarity connection. Final energy \(U_f = \frac{1}{2}(2C)V^2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{C}{4}(V_1^2 + V_2^2 + 2V_1V_2)\) . Decrease in energy \(Delta U = U_i - U_f = \frac{1}{4}C(V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4}C(V_1 - V_2)^2\).

Concept: Work done by external agent in removing dielectric from capacitor connected to battery. Formula: (W_{\text{ext}} = \Delta U - W_{\text{cell}}\), where (W_{\text{cell}} = E\Delta Q\). Solution: Initial stored energy (U_1 = \frac{1}{2} kCE^2\) and charge (Q_1 = kCE\). Final stored energy (U_2 = \frac{1}{2} CE^2\) and charge (Q_2 = CE\). Change in stored energy (Delta U = U_2 - U_1 = -\frac{1}{2} CE^2(k-1)\). Charge returned to cell (Delta Q = Q_1 - Q_2 = CE(k-1)\). Work done by cell (W_{\text{cell}} = E (Q_2 - Q_1) = -E^2C(k-1)\). Work done by external agent (W_{\text{ext}} = \Delta U - W_{\text{cell}} = -\frac{1}{2} CE^2(k-1) - (-E^2C(k-1)) = \frac{1}{2} E^2C(k-1)\).

In series, the equivalent capacitance is \(C_{text{eq}} = \frac{C}{2}\). Since the voltage is shared equally between the two identical capacitors, the maximum safe voltage across the combination is \(2V\).