NEET Combination of Capacitors MCQs & PYQs

Let's verify and calculate the correct answer step-by-step:

Given Data:

• Capacitances:
  $C_1 = 2 \, \mu\text{F}, C_2 = 3 \, \mu\text{F}, C_3 = 5 \, \mu\text{F}$ 
• Maximum voltages:
  $V_1 = 3 \, \text{V}, V_2 = 2 \, \text{V}, V_3 = 1 \, \text{V}$ 

Step 1: Maximum charge each capacitor can store:

$$Q_1 = C_1 \cdot V_1 = 2 \cdot 3 = 6 \, \mu\text{C}$$

$$Q_2 = C_2 \cdot V_2 = 3 \cdot 2 = 6 \, \mu\text{C}$$

$$Q_3 = C_3 \cdot V_3 = 5 \cdot 1 = 5 \, \mu\text{C}$$

The capacitor with the minimum charge capacity limits the system. Here,

$Q_{\text{max}} = 5 \, \mu\text{C}$

, dictated by

$C_3$

.

Step 2: Voltage distribution across each capacitor:

In series, charge

$Q$

is the same on all capacitors, and the voltage across each capacitor is:

$$V_1 = \frac{Q}{C_1}, \quad V_2 = \frac{Q}{C_2}, \quad V_3 = \frac{Q}{C_3}$$

Total voltage across the series combination:

$$V_{\text{total}} = V_1 + V_2 + V_3$$

Substitute

$Q = 5 \, \mu\text{C}$

:

$$V_1 = \frac{5}{2} = 2.5 \, \text{V}, \quad V_2 = \frac{5}{3} \approx 1.67 \, \text{V}, \quad V_3 = \frac{5}{5} = 1 \, \text{V}$$

Step 3: Total voltage:

$$V_{\text{total}} = V_1 + V_2 + V_3 = 2.5 + 1.67 + 1 = \frac{15}{6} + \frac{10}{6} + \frac{6}{6} = \frac{31}{6} \, \text{V}.$$

Final Answer:

The maximum voltage the series combination can withstand is:

$$\boxed{\frac{31}{6} \, \text{V}} \approx 5.17 \, \text{V}.$$

To find the equivalent capacitance for this cubical capacitor network, where each edge of the cube has a capacitance

$C$

, here’s the shortest solution:

Step-by-Step:

1. Symmetry analysis:
   • By symmetry, all corners of the cube can be grouped into equivalent potential nodes.
   • The cube's symmetry allows reduction to a simpler circuit.
2. Key nodes:
   • Node
     $A$ 

     is connected to one corner of the cube.

   • Node
     $B$ 

     is connected to the diagonally opposite corner.

3. Effective connections:
   • Due to symmetry, three capacitors are effectively in parallel between
     $A$ 

     and an intermediate point.

   • Similarly, three capacitors are effectively in parallel between
     $B$ 

     and the same intermediate point.

   • Two capacitors remain directly between
     $A$ 

     and $B$ 

     .

4. Simplification:
   • The three parallel capacitors at each node result in:
     $$C_{\text{parallel}} = 3C$$ 
   • The equivalent circuit becomes two
     $3C$ 

     capacitors in series with a $2C$ 

     capacitor: $$\text{Series combination:} \quad \frac{1}{C_{\text{eq}}} = \frac{1}{3C} + \frac{1}{3C} + \frac{1}{2C}$$ 

5. Calculation:
   • Combine series:
     $$\frac{1}{C_{\text{eq}}} = \frac{2}{3C} + \frac{1}{2C} = \frac{4}{6C} + \frac{3}{6C} = \frac{7}{6C}$$ 
   • Invert to find
     $C_{\text{eq}}$ 

     : $$C_{\text{eq}} = \frac{6C}{7} \times 2 = \frac{12C}{7}$$ 

Thus, the equivalent capacitance is:

$$C_{\text{eq}} = \frac{12C}{7}$$

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 \(V_{max} = \frac{Q_{max}}{C_{eq}} = \frac{5 \mu\text{C}}{(30/31) \mu\text{F}} = \frac{5 \times 31}{30} = \frac{31}{6} \text{ Volts}\).