(a)  A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

(b) A parallel plate capacitor of capacitance C is charged to a potential V by a battery.Without disconnecting the battery, the distance between the plates is tripled and a dielectric medium of k = 10 is introduced between the plates is tripled and a dielectric medium of k = 10 is introduced between the plates of the capacitor.
Explain giving reasons, how will the following be affected: 

(i) capacitance of the capacitor 

(ii) charge on the capacitor, and

(iii) energy density of the capacitor

                                                             OR

(a) Describe the working of Light Emitting Diodes (LEDs). Which semiconductors are preferred to make LEDs and why? Give two advantages of using LEDs over conventional incandescent low power lamps. 

(b) If each diode in figure has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the currents ${\mathrm{I}}_1,{\mathrm{I}}_2,{\mathrm{I}}_3 \mathrm{and} {\mathrm{I}}_4?$

Part Q (a): Energy Distribution in Capacitor Systems

Problem Analysis

When a charged capacitor (capacitance C, potential V) is connected to an identical uncharged capacitor, we need to find the energy ratio.

Initial Conditions

• Initial energy stored: U₁ = ½CV²
• Initial charge: Q = CV

After Connection

When the capacitors are connected in parallel:

• Total charge remains constant: Q_total = CV
• This charge distributes equally between both capacitors
• Each capacitor gets charge: Q/2 = CV/2
• Common potential across both: V_final = Q_total/(2C) = V/2

Final Energy Calculation

Energy in combined system:

• U₂ = 2 × [½C(V/2)²] = 2 × [½C × V²/4] = CV²/4

Energy Ratio

Ratio = U₂/U₁ = (CV²/4)/(½CV²) = 1/2

Key Insight: Half the initial energy is lost as heat during charge redistribution due to the transient current flow in the connecting wires.

Part Q (b): Effects of Modifying Capacitor Configuration

Initial Setup

• Capacitance: C
• Charged to potential: V (by battery, still connected)
• Changes: Distance tripled (d → 3d) and dielectric with k = 10 inserted

(i) Effect on Capacitance

Initial capacitance: C = ε₀A/d

After modifications: C' = kε₀A/(3d) = 10ε₀A/(3d)

New capacitance: C' = (10/3)C ≈ 3.33C

Result: Capacitance increases by a factor of 10/3 because the dielectric effect (×10) dominates over the distance increase (÷3).

(ii) Effect on Charge

Since the battery remains connected, the potential V stays constant.

Initial charge: Q = CV

New charge: Q' = C'V = (10/3)CV = (10/3)Q

Result: Charge increases by a factor of 10/3 because the battery supplies additional charge to maintain constant voltage across the increased capacitance.

(iii) Effect on Energy Density

Energy density = Energy per unit volume = u = ½ε₀E²

Initial electric field: E = V/d

New electric field: E' = V/(3d) = E/3

Initial energy density: u = ½ε₀(V/d)²

New energy density inside dielectric: u' = ½(kε₀)(V/3d)² = ½(10ε₀)(V²/9d²)

Comparing: u'/u = [10ε₀V²/(9d²)]/[ε₀V²/d²] = 10/9

Result: Energy density increases by a factor of 10/9. Though the electric field decreases by 1/3, the dielectric constant of 10 more than compensates, leading to a net increase.

LED Questions

Part (a): Working Principle and Advantages of LEDs

Working Principle of LEDs

Light Emitting Diodes are semiconductor devices that emit light through electroluminescence when forward-biased. The process involves:

1. P-N Junction Formation: An LED consists of a p-type and n-type semiconductor junction
2. Forward Bias Application: When voltage is applied with positive terminal to p-side, electrons from n-region and holes from p-region move toward the junction
3. Recombination Process: Electrons and holes recombine at the junction, releasing energy
4. Photon Emission: The energy released appears as photons (light) with energy approximately equal to the band gap energy: E = hν = Eg

Preferred Semiconductors for LEDs

Direct band gap semiconductors are preferred, including:

• Gallium Arsenide (GaAs): Infrared LEDs
• Gallium Phosphide (GaP): Red and green LEDs
• Gallium Nitride (GaN): Blue, white, and UV LEDs
• Indium Gallium Nitride (InGaN): High-brightness blue and white LEDs
• Aluminum Gallium Indium Phosphide (AlGaInP): High-brightness red, orange, and yellow LEDs

Why Direct Band Gap Materials? In direct band gap semiconductors, electron-hole recombination directly produces photons with high efficiency (radiative recombination). Indirect band gap materials (like silicon) primarily release energy as heat (phonons), making them inefficient for light emission.

Two Major Advantages of LEDs over Incandescent Lamps

1. Energy Efficiency: LEDs convert 80-90% of electrical energy into light, while incandescent bulbs convert only 5-10%, with the rest wasted as heat. This results in significantly lower power consumption for the same light output.
2. Longevity: LEDs have operational lifetimes of 25,000-50,000 hours compared to 1,000-2,000 hours for incandescent bulbs, reducing replacement costs and maintenance.

Additional advantages include instant illumination, durability, compact size, and absence of toxic materials like mercury.

Part (b): Circuit Analysis with Diodes

Circuit Configuration Analysis

Given specifications:

• Forward bias resistance: 25 Ω per diode
• Reverse bias resistance: ∞ (infinite) per diode
• Supply voltage: 5 V
• Each branch has a 125 Ω resistor in series with a diode

Current Determination

Since diodes have infinite reverse resistance, only forward-biased diodes will conduct current.

Analyzing the circuit from the diagram:

• Points on the left (A, C, E, G) are at higher potential than points on the right (B, D, F, H)
• All four diodes are forward-biased
• All branches conduct current in parallel

For each branch: Total resistance per branch = 125 Ω + 25 Ω = 150 Ω

Current through each diode: I = V/R_total = 5V/150Ω = 1/30 A ≈ 0.0333 A = 33.3 mA

Final Answer

I₁ = I₂ = I₃ = I₄ = 1/30 A = 0.0333 A (or 33.3 mA)

All currents are equal because all diodes are identically forward-biased with the same resistance values in each parallel branch.

Summary

This problem set covers fundamental concepts in capacitor energy storage, electrostatics, and semiconductor devices, demonstrating practical applications in energy systems and modern lighting technology.