Axiom Physics | Class 12 - EMF, Internal Resistance & Cells

⚡ 01. The Spark (Analogy)

Socho ek water pump hai jo 100 litre/min paani phekne ki theoretical capacity rakhta hai (EMF $\mathcal{E}$). Lekin pump ke andar ki pipe aur impeller me friction hota hai (Internal Resistance $r$). Jab pump se paani bahar deliver hona shuru hota hai, toh is internal friction ke chalte kuch pressure andar hi drop ho jata hai ($Ir$). Isliye bahar jo usable output milta hai (Terminal Voltage $V$), woh theoretical capacity se hamesha kam hota hai: $V = \mathcal{E} - Ir$[cite: 13]! Lekin agar aap kisi bahari source se battery ko force recharge kar rahe hon, toh battery ko internal resistance ke against push karna padta hai, aur tab $V = \mathcal{E} + Ir$ ho jata hai!

📐 02. The Core Law & Formulae

1. EMF & Terminal Voltage ($V$)

Relation during discharging (driving current through external load $R$)[cite: 13]:

$$I = \frac{\mathcal{E}}{R + r}, \quad V = \mathcal{E} - I r = I R$$

Internal resistance formula: $r = \left(\frac{\mathcal{E} - V}{V}\right)R = \left(\frac{\mathcal{E}}{V} - 1\right)R$[cite: 13].

2. Grouping of Non-Identical Cells

Equivalent EMF and internal resistance in series and parallel[cite: 13]:

$$\text{Series: } \mathcal{E}_{\text{eq}} = \mathcal{E}_1 + \mathcal{E}_2, \quad r_{\text{eq}} = r_1 + r_2$$ $$\text{Parallel: } \mathcal{E}_{\text{eq}} = \frac{\frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}, \quad \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}$$

Maximum power delivered to load occurs when load matches source: $R = r$ ($P_{\max} = \frac{\mathcal{E}^2}{4r}$).

03. Interactive Visualizer: Real Battery Circuit Bench

External Load ($R$) 4.0 Ω

Cell EMF ($\mathcal{E}$) 10.0 V

Internal Resistance ($r$) 1.0 Ω

Terminal Voltage ($V = \mathcal{E} - Ir$):

8.00 V

Circuit Current ($I$):

2.00 A

Power Delivered to Load ($P = I^2 R$):

16.00 W

📝 04. Stepwise Derivations Blueprint[cite: 13]

Derivation A

Equivalent EMF & Internal Resistance of Two Cells in Series[cite: 13]

1. Consider two cells of EMFs $\mathcal{E}_1, \mathcal{E}_2$ and internal resistances $r_1, r_2$ connected in series between terminals $A$ and $C$, with common node $B$[cite: 13].

2. Potential differences across each cell[cite: 13]:

$$V_{AB} = V_A - V_B = \mathcal{E}_1 - I r_1$$ $$V_{BC} = V_B - V_C = \mathcal{E}_2 - I r_2$$

3. Net terminal potential difference across the combination $V_{AC} = V_A - V_C$[cite: 13]:

$$V_{AC} = (V_A - V_B) + (V_B - V_C) = (\mathcal{E}_1 - I r_1) + (\mathcal{E}_2 - I r_2)$$ $$V_{AC} = (\mathcal{E}_1 + \mathcal{E}_2) - I (r_1 + r_2)$$

4. Comparing with equivalent cell equation $V_{AC} = \mathcal{E}_{\text{eq}} - I r_{\text{eq}}$[cite: 13]:

$$\mathcal{E}_{\text{eq}} = \mathcal{E}_1 + \mathcal{E}_2, \quad r_{\text{eq}} = r_1 + r_2 \text{[cite: 13]}$$

Opposition Polarity: If cell 2 is reversed (positive to positive), $\mathcal{E}_{\text{eq}} = \mathcal{E}_1 - \mathcal{E}_2$ (assuming $\mathcal{E}_1 > \mathcal{E}_2$), but internal resistances still add: $r_{\text{eq}} = r_1 + r_2$.

Derivation B

Equivalent EMF & Internal Resistance of Two Cells in Parallel[cite: 13]

1. Consider two cells with parameters $(\mathcal{E}_1, r_1)$ and $(\mathcal{E}_2, r_2)$ connected in parallel between nodes $B_1$ and $B_2$[cite: 13].

2. Let main current be $I = I_1 + I_2$[cite: 13]. Both branches share the identical terminal potential difference $V = V_{B1} - V_{B2}$[cite: 13]:

$$V = \mathcal{E}_1 - I_1 r_1 \implies I_1 = \frac{\mathcal{E}_1 - V}{r_1}$$ $$V = \mathcal{E}_2 - I_2 r_2 \implies I_2 = \frac{\mathcal{E}_2 - V}{r_2}$$

3. Total current $I$[cite: 13]:

$$I = I_1 + I_2 = \left(\frac{\mathcal{E}_1 - V}{r_1}\right) + \left(\frac{\mathcal{E}_2 - V}{r_2}\right) = \left(\frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2}\right) - V\left(\frac{1}{r_1} + \frac{1}{r_2}\right)$$

4. Rearranging for terminal potential $V$[cite: 13]:

$$V\left(\frac{r_1 + r_2}{r_1 r_2}\right) = \left(\frac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 r_2}\right) - I$$ $$V = \left(\frac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 + r_2}\right) - I \left(\frac{r_1 r_2}{r_1 + r_2}\right)$$

5. Comparing with $V = \mathcal{E}_{\text{eq}} - I r_{\text{eq}}$[cite: 13]:

$$\mathcal{E}_{\text{eq}} = \frac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 + r_2} = \frac{\sum \mathcal{E}_i / r_i}{\sum 1 / r_i}, \quad \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} \text{[cite: 13]}$$

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Can Terminal Voltage $V$ exceed EMF $\mathcal{E}$?

YES! Students blindly memorize $V < \mathcal{E}$[cite: 13].
• Discharging (Normal use): Current leaves positive terminal $\implies \mathbf{V = \mathcal{E} - Ir < \mathcal{E}}$[cite: 13].
• Open Circuit ($I = 0$): $\mathbf{V = \mathcal{E}}$[cite: 13].
• Charging (Battery Charger active): Current enters positive terminal $\implies \mathbf{V = \mathcal{E} + Ir > \mathcal{E}}$!

Trap 2: The Reversed Cell Trap in Series

If out of $n$ identical cells, $m$ cells are connected in reverse polarity:
Net assisting cells $= n - m$, opposing cells $= m$.
Net EMF: $\mathcal{E}_{\text{net}} = (n - m)\mathcal{E} - m\mathcal{E} = \mathbf{(n - 2m)\mathcal{E}}$.
Total internal resistance: **Remains $n r$** (internal resistances add regardless of polarity!).

Trap 3: Maximum Power Transfer vs Efficiency

Power delivered to load is maximum when $R = r$ ($P_{\max} = \frac{\mathcal{E}^2}{4r}$). But at this point, terminal voltage is $V = \mathcal{E}/2$ and electrical efficiency $\eta = \frac{V}{\mathcal{E}} = \mathbf{50\%}$ only. Maximum power condition is NOT maximum efficiency condition!

Trap 4: Short Circuit Current ($R = 0$)

When terminals of a battery are directly shorted with zero load resistance ($R = 0$), $I_{\text{sc}} = \frac{\mathcal{E}}{r}$ and terminal voltage drops to exactly $V = 0$[cite: 13]. All power is dissipated internally as destructive heat ($P_{\text{loss}} = \mathcal{E}^2 / r$).

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A battery of EMF $10\text{ V}$ and internal resistance $3\text{ }\Omega$ is connected to a resistor. If the current in the circuit is $0.5\text{ A}$, find (i) the resistance of the resistor, (ii) terminal voltage of the battery when the circuit is closed[cite: 13].

Solution:

1. Circuit equation: $I = \frac{\mathcal{E}}{R + r} \implies 0.5 = \frac{10}{R + 3}$[cite: 13]:

$$R + 3 = \frac{10}{0.5} = 20 \implies R = 17\text{ }\Omega$$

2. Terminal voltage $V$ during discharge[cite: 13]:

$$V = \mathcal{E} - I r = 10 - (0.5 \times 3) = 10 - 1.5 = 8.5\text{ Volts}$$

Cross Check: $V = IR = 0.5 \times 17 = 8.5\text{ V}$.

JEE/NEET Competitive Trap

Problem: Two cells of EMFs $2\text{ V}$ and $1\text{ V}$ with internal resistances $1\text{ }\Omega$ and $2\text{ }\Omega$ respectively are connected in parallel with like terminals together. The combination is connected to an external resistor of $10\text{ }\Omega$. Find the current through each cell and the terminal potential difference[cite: 13].

Solution:

1. Equivalent EMF and internal resistance[cite: 13]:

$$r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{1 \times 2}{1 + 2} = \frac{2}{3}\text{ }\Omega$$ $$\mathcal{E}_{\text{eq}} = \frac{\frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{2}{1} + \frac{1}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{2.5}{1.5} = \frac{5}{3}\text{ Volts}$$

2. Total current through external load $R = 10\text{ }\Omega$[cite: 13]:

$$I = \frac{\mathcal{E}_{\text{eq}}}{R + r_{\text{eq}}} = \frac{5/3}{10 + 2/3} = \frac{5/3}{32/3} = \frac{5}{32}\text{ A} \approx 0.156\text{ A}$$

3. Terminal voltage across the load[cite: 13]:

$$V = I R = \frac{5}{32} \times 10 = \frac{50}{32} = \frac{25}{16}\text{ V} \approx 1.5625\text{ Volts}$$

4. Individual currents: $I_1 = \frac{\mathcal{E}_1 - V}{r_1} = \frac{2 - 1.5625}{1} = 0.4375\text{ A}$, $I_2 = \frac{\mathcal{E}_2 - V}{r_2} = \frac{1 - 1.5625}{2} = -0.281\text{ A}$.

Trap Alert: Notice $I_2$ is negative! Because cell 1 EMF ($2\text{ V}$) is significantly higher than cell 2 EMF ($1\text{ V}$), cell 1 is actually discharging while back-charging cell 2[cite: 13]!

🗂️ 07. Quick Revision Flashcard Core Formulae

Discharging Cell

$V = \mathcal{E} - Ir$[cite: 13]

Charging Cell

$V = \mathcal{E} + Ir$

Parallel Equivalent

$\mathcal{E}_{\text{eq}} = \frac{\mathcal{E}_1 r_2 + \mathcal{E}_2 r_1}{r_1 + r_2}$[cite: 13]

Max Power Condition

$R = r \implies P_{\max} = \frac{\mathcal{E}^2}{4r}$

EMF, Internal Resistance & Combination of Cells