Axiom Physics | Class 12 - Kirchhoff's Rules & Wheatstone Bridge

⚡ 01. The Spark (Analogy)

Socho ek busy road intersection ko: jitni gaadiyan junction par enter karti hain, utni hi bahar nikalni padti hain—koi gaadi hawa me gayab nahi ho sakti (Kirchhoff's Junction Rule / Conservation of Charge). Aur agar aap kisi amusement park ke roller coaster track par ek complete round lagakar wapas apni starting seat par aate hain, toh pure ride ke dauran total gain in altitude aur drop in altitude ka net sum zero hota hai (Kirchhoff's Loop Rule / Conservation of Energy). Jab inhi do rules ko ek diamond network me apply karte hain aur opposite sides ka potential drop equal ho jata hai, toh central bridge branch me current zero ho jata hai—isi ko Balanced Wheatstone Bridge kehte hain!

📐 02. The Core Law & Formulae

1. Kirchhoff's First & Second Laws (KCL & KVL)

Junction rule and closed loop potential summation:

$$\sum I_{\text{in}} = \sum I_{\text{out}} \implies \sum I = 0 \quad (\text{Junction Rule})$$ $$\oint \Delta V = 0 \implies \sum \mathcal{E} + \sum (-IR) = 0 \quad (\text{Loop Rule})$$

Physical origin: KCL is based on Conservation of Charge; KVL is based on Conservation of Energy.

2. Wheatstone Bridge Condition

Null deflection condition in the galvanometer branch ($I_g = 0$):

$$V_B = V_D \iff \frac{P}{Q} = \frac{R}{S}$$

Where $P, Q, R, S$ are the four resistance arms. When balanced, the bridge has maximum sensitivity when all four resistors are of equal order of magnitude.

03. Interactive Visualizer: Wheatstone Bridge Balancing Engine

Arm P 10.0 Ω

Arm Q 20.0 Ω

Arm R 15.0 Ω

Arm S (Variable) 25.0 Ω

Ratio P / Q:

0.500

Ratio R / S:

0.600

Bridge Deflection ($I_g$):

+14.2 mA

Bridge State:

UNBALANCED

📝 04. Stepwise Derivations Blueprint

Derivation

Deduction of Wheatstone Bridge Balancing Condition Using Kirchhoff's Laws

1. Consider four resistors $P, Q, R, S$ forming closed loop $ABDA$ and $BCDB$. Battery of EMF $\mathcal{E}$ is connected between $A$ and $C$, and galvanometer of resistance $G$ is connected between $B$ and $D$.

2. Let main current entering node $A$ be $I$. At node $A$, current splits into $I_1$ through arm $AB$ (resistor $P$) and $I_2$ through arm $AD$ (resistor $R$).

3. At node $B$, current $I_g$ flows into the galvanometer. Remaining current through arm $BC$ (resistor $Q$) is $(I_1 - I_g)$.

4. At node $D$, current $I_2$ and $I_g$ join together, so current flowing through arm $DC$ (resistor $S$) is $(I_2 + I_g)$.

• Apply Kirchhoff's Loop Rule (KVL) to Loop 1 ($ABDA$):

$$-I_1 P - I_g G + I_2 R = 0 \implies I_1 P + I_g G = I_2 R \quad \text{--- [Eq. 1]}$$

• Apply Kirchhoff's Loop Rule (KVL) to Loop 2 ($BCDB$):

$$-(I_1 - I_g)Q + (I_2 + I_g)S + I_g G = 0 \quad \text{--- [Eq. 2]}$$

• Balanced Condition (Null Deflection in Galvanometer):

When the bridge is balanced, potential at node $B$ equals potential at node $D$ ($V_B = V_D$). Thus, no current flows through the galvanometer: $I_g = 0$.

Substitute $I_g = 0$ into Eq. 1:

$$I_1 P = I_2 R \quad \text{--- [Eq. 3]}$$

Substitute $I_g = 0$ into Eq. 2:

$$I_1 Q = I_2 S \quad \text{--- [Eq. 4]}$$

Dividing Eq. 3 by Eq. 4:

$$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \implies \frac{P}{Q} = \frac{R}{S}$$

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Interchanging Battery and Galvanometer

*Board Question:* What happens to the balance condition if the battery and galvanometer are interchanged?
*Answer:* The balance condition REMAINS UNCHANGED! If $P/Q = R/S$, interchanging source and detector simply swaps conjugate arms ($P/R = Q/S$). The galvanometer still shows zero deflection.

Trap 2: KVL Sign Convention Mishaps

• Moving across a resistor *in the direction of current* $\implies$ Potential drop: $\mathbf{-IR}$.
• Moving across a resistor *opposite to current* $\implies$ Potential gain: $\mathbf{+IR}$.
• Moving through a cell from negative to positive terminal $\implies \mathbf{+\mathcal{E}}$.
• Moving from positive to negative terminal $\implies \mathbf{-\mathcal{E}}$.

Trap 3: Maximum Bridge Sensitivity

A Wheatstone bridge is most sensitive (shows maximum noticeable needle deflection for the smallest change in resistance) when all four arms have nearly equal resistances ($P \approx Q \approx R \approx S$). If one arm is $1\text{ }\Omega$ and another is $10{,}000\text{ }\Omega$, sensitivity drops dramatically.

Trap 4: Resistance of Central Galvanometer Arm at Balance

When the bridge is balanced ($I_g = 0$), the central galvanometer branch carries no current. Therefore, it can be completely removed from equivalent resistance calculations! The network simplifies to $(P+Q)$ in parallel with $(R+S)$: $R_{\text{eq}} = \frac{(P+Q)(R+S)}{(P+Q) + (R+S)}$.

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: The four arms of a Wheatstone bridge have the following resistances: $AB = 100\text{ }\Omega$, $BC = 10\text{ }\Omega$, $CD = 5\text{ }\Omega$, and $DA = 60\text{ }\Omega$. A galvanometer of resistance $15\text{ }\Omega$ is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10\text{ V}$ is maintained across $AC$.

Solution:

1. Check bridge balance: $\frac{P}{Q} = \frac{100}{10} = 10$; $\frac{R}{S} = \frac{60}{5} = 12$. Since $10 \ne 12$, bridge is unbalanced ($I_g \ne 0$).

2. Apply KVL to Loop 1 ($ABDA$):

$$-100 I_1 - 15 I_g + 60 I_2 = 0 \implies 20 I_1 + 3 I_g - 12 I_2 = 0 \quad \text{--- [1]}$$

3. Apply KVL to Loop 2 ($BCDB$):

$$-10(I_1 - I_g) + 15 I_g + 5(I_2 + I_g) = 0 \implies -10 I_1 + 30 I_g + 5 I_2 = 0 \implies 2 I_1 - 6 I_g - I_2 = 0 \quad \text{--- [2]}$$

4. From Eq. [2]: $I_2 = 2 I_1 - 6 I_g$. Substitute $I_2$ into Eq. [1]:

$$20 I_1 + 3 I_g - 12(2 I_1 - 6 I_g) = 0 \implies -4 I_1 + 75 I_g = 0 \implies I_1 = \frac{75}{4} I_g = 18.75 I_g$$

5. Substitute $I_1$ to get $I_2$: $I_2 = 2(18.75 I_g) - 6 I_g = 31.5 I_g$.

6. Apply KVL along outer path through source ($AC$): $100 I_1 + 10(I_1 - I_g) = 10 \implies 110 I_1 - 10 I_g = 10$.

Substitute $I_1 = 18.75 I_g$:

$$110(18.75 I_g) - 10 I_g = 10 \implies 2062.5 I_g - 10 I_g = 10 \implies 2052.5 I_g = 10$$ $$I_g = \frac{10}{2052.5} \approx 4.87 \times 10^{-3}\text{ A} = 4.87\text{ mA}$$

JEE/NEET Competitive Trap

Problem: In a Wheatstone bridge, $P = 2\text{ }\Omega, Q = 3\text{ }\Omega, R = 6\text{ }\Omega$, and $S = 8\text{ }\Omega$. What resistance $X$ must be connected in parallel across $S$ to balance the bridge?

Solution:

1. For the bridge to be balanced, the effective resistance in the fourth arm ($S'$) must satisfy:

$$\frac{P}{Q} = \frac{R}{S'} \implies \frac{2}{3} = \frac{6}{S'} \implies S' = \frac{6 \times 3}{2} = 9\text{ }\Omega$$

2. But the existing resistance $S = 8\text{ }\Omega$. Notice that connecting a resistor $X$ in parallel can ONLY decrease the equivalent resistance below $8\text{ }\Omega$ ($S_{\text{parallel}} < 8\text{ }\Omega$). It can NEVER increase resistance to $9\text{ }\Omega$!

$$\text{Verdict: No physical positive resistance } X \text{ can balance the bridge in parallel!}$$

Exam Trap Note: To balance this bridge, a resistance of $1\text{ }\Omega$ must be connected in series with $S$ ($8 + 1 = 9\text{ }\Omega$), not in parallel. Examiners deliberately set this trap to catch blind formula substitution!

🗂️ 07. Quick Revision Flashcard Core Formulae

Kirchhoff Junction (KCL)

$\sum I = 0$ (Charge Cons.)

Kirchhoff Loop (KVL)

$\sum \Delta V = 0$ (Energy Cons.)

Wheatstone Balance

$\frac{P}{Q} = \frac{R}{S} \iff I_g = 0$

Max Sensitivity

$P \approx Q \approx R \approx S$

Kirchhoff's Rules & The Wheatstone Bridge