Axiom Physics | Class 12 - Capacitor Networks & Energy

⚡ 01. The Spark (Analogy)

Socho overhead water tanks ko: agar aap 2 tanks ko side-by-side (parallel) connect karte hain, toh unka total cross-section badh jata hai aur woh double paani store kar sakte hain[cite: 13]. Lekin agar unhe vertical pipe se series mein connect karte hain, toh pressure divide ho jata hai aur overall flow restrict hota hai[cite: 13]. Theek waise hi, Parallel Capacitors directly add hote hain ($C_{\text{eq}} = C_1 + C_2$), jabki Series Capacitors reciprocal add hote hain ($1/C_{\text{eq}} = 1/C_1 + 1/C_2$)[cite: 13]. Aur jab battery disconnected rehti hai, toh charge lock ho jata hai; jab battery connected rehti hai, toh voltage lock rehti hai!

📐 02. The Core Law & Formulae

Series & Parallel Grouping

Series (same charge $Q$) vs Parallel (same voltage $V$)[cite: 13]:

$$\text{Series: } \frac{1}{C_s} = \sum_{i=1}^n \frac{1}{C_i}, \quad \text{Parallel: } C_p = \sum_{i=1}^n C_i$$

Notice: $C_s < C_{\min}$ always, while $C_p > C_{\max}$ always[cite: 13].

Stored Energy & Energy Density

Electrostatic energy stored in the electric field between plates[cite: 13]:

$$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV, \quad u_E = \frac{1}{2}\varepsilon_0 E^2$$

SI unit of energy density $u_E$: $\text{J/m}^3$. Stored in the volume $Ad$ of space.

03. Interactive Visualizer: Circuit Bench & Battery State Engine

Capacitor $C_1$ 4.0 μF

Capacitor $C_2$ 6.0 μF

Supply Voltage ($V_0$) 12 V

Equivalent Capacitance ($C_{\text{eq}}$):

2.40 μF

Total Circuit Charge ($Q_{\text{total}}$):

28.8 μC

Total Stored Energy ($U = \frac{1}{2}C_{\text{eq}}V^2$):

172.8 μJ

📝 04. Stepwise Derivations Blueprint[cite: 13]

Derivation A

Capacitors in Series Combination[cite: 13]

1. Connect $C_1, C_2, \dots, C_n$ end-to-end across voltage $V$. By charge conservation, every capacitor acquires the same charge $Q$[cite: 13].

2. Total voltage drop $V$ equals the sum of potential drops across individual capacitors[cite: 13]:

$$V = V_1 + V_2 + \dots + V_n$$

3. Substitute $V_i = \frac{Q}{C_i}$ and $V = \frac{Q}{C_s}$[cite: 13]:

$$\frac{Q}{C_s} = \frac{Q}{C_1} + \frac{Q}{C_2} + \dots + \frac{Q}{C_n}$$

4. Canceling $Q$ yields the reciprocal addition law[cite: 13]:

$$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}$$

Derivation B

Capacitors in Parallel Combination[cite: 13]

1. Connect $C_1, C_2, \dots, C_n$ across common terminal nodes. Each capacitor experiences the same potential difference $V$[cite: 13].

2. Total charge supplied by source $Q = Q_1 + Q_2 + \dots + Q_n$[cite: 13].

3. Substitute $Q_i = C_i V$ and $Q = C_p V$[cite: 13]:

$$C_p V = C_1 V + C_2 V + \dots + C_n V$$

4. Canceling common potential $V$ yields direct sum[cite: 13]:

$$C_p = C_1 + C_2 + \dots + C_n$$

Derivation C

Common Potential & Energy Loss in Parallel Sharing

1. When two capacitors $C_1$ (charged to $V_1$) and $C_2$ (charged to $V_2$) are connected in parallel, charge flows until common potential $V_c$ is reached:

$$V_c = \frac{Q_1 + Q_2}{C_1 + C_2} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$

2. Electrostatic energy loss ($\Delta U = U_i - U_f$) dissipated strictly as heat and spark radiation:

$$\Delta U = \frac{1}{2}\frac{C_1 C_2}{C_1 + C_2}(V_1 - V_2)^2 \ge 0$$

⚠️ 05. Red Alert Exam Trap: The "Battery State" Matrix

Board aur JEE dono ka sabse favorite trap: Dielectric slab insert karte waqt battery connected hai ya remove kar di gayi hai?

Physical Quantity	Case I: Battery Remains Connected	Case II: Battery Disconnected
Capacitance ($C$)	Increases ($K C_0$)[cite: 13]	Increases ($K C_0$)[cite: 13]
Potential Difference ($V$)	CONSTANT ($V_0$)	Decreases ($V_0 / K$)[cite: 13]
Charge on Plates ($Q$)	Increases ($K Q_0$)	CONSTANT ($Q_0$)
Electric Field ($E$)	CONSTANT ($E_0 = V_0/d$)	Decreases ($E_0 / K$)[cite: 13]
Stored Energy ($U$)	Increases ($K U_0$) [Battery does work]	Decreases ($U_0 / K$) [Field pulls slab in]

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A $900\text{ pF}$ capacitor is charged by a $100\text{ V}$ battery. (a) How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another uncharged $900\text{ pF}$ capacitor. How much electrostatic energy is stored in the system now[cite: 13]?

Solution:

1. Initial energy stored ($U_i$)[cite: 13]:

$$U_i = \frac{1}{2} C V^2 = \frac{1}{2} \times (900 \times 10^{-12}\text{ F}) \times (100\text{ V})^2 = 4.5 \times 10^{-6}\text{ Joules}$$

2. Disconnected and connected in parallel with identical uncharged $C_2 = 900\text{ pF}$[cite: 13]:

$$V_c = \frac{C_1 V_1 + 0}{C_1 + C_2} = \frac{900 \times 100}{900 + 900} = 50\text{ Volts}$$

3. Total final energy ($U_f$):

$$U_f = \frac{1}{2}(C_1 + C_2) V_c^2 = \frac{1}{2} \times (1800 \times 10^{-12}) \times (50)^2 = 2.25 \times 10^{-6}\text{ Joules}$$

$$\text{Energy Loss } \Delta U = U_i - U_f = 2.25 \times 10^{-6}\text{ J (50% energy lost as heat/radiation)}$$

JEE/NEET Competitive Trap

Problem: A capacitor $C$ is charged to voltage $V_0$. The charging battery is removed, and a dielectric slab of constant $K = 4$ is slipped between the plates. What is the work done by the electrostatic field in pulling the dielectric slab inside?

Solution:

1. Battery is removed $\implies$ Charge $Q = C V_0$ remains strictly constant.

2. Initial energy: $U_i = \frac{Q^2}{2C}$[cite: 13].

3. Final capacitance becomes $K C = 4C$[cite: 13]. Final energy: $U_f = \frac{Q^2}{2(4C)} = \frac{U_i}{4}$[cite: 13].

4. By work-energy theorem ($W_{\text{field}} = -\Delta U = U_i - U_f$):

$$W_{\text{field}} = U_i - \frac{U_i}{4} = \frac{3}{4} U_i = \frac{3}{8} C V_0^2$$

Exam Trap Note: Because field does positive work, the dielectric experiences an attractive suction force pulling it into the capacitor plates[cite: 13].

🗂️ 07. Quick Revision Flashcard Core Formulae

Series Formula

$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}$[cite: 13]

Parallel Formula

$C_p = C_1 + C_2$[cite: 13]

Energy Density

$u_E = \frac{1}{2}\varepsilon_0 E^2$

Energy Loss Formula