Class 11> Unit # 10: DC Circuits > Wheat-Stone Bridge

Wheatstone Bridge - Talha's Physics Academy

Talha's Physics Academy

Wheatstone Bridge & Balanced Condition

Video Lecture

Watch the complete video lecture below to understand the Wheatstone Bridge configuration and step-by-step mathematical derivation for its balanced condition.

Wheatstone Bridge

"When four resistors are connected in such a way that a potential difference source is connected between two common junctions, while a sensitive galvanometer is connected between the other two common junctions, this combination is called a Wheatstone bridge."

Circuit Explanation

Consider four resistors connected in a bridge network between common junctions $a$, $b$, $c$, and $d$:

  • A voltage source (battery) and key $K_1$ are connected between junctions $a$ and $c$.
  • A galvanometer of resistance $G$ and key $K_2$ are connected between junctions $b$ and $d$.

When keys $K_1$ and $K_2$ are closed, current flows through the circuit. The magnitudes of the four resistors are adjusted until no current flows through the galvanometer ($I_g = 0$). Under this condition, the bridge is referred to as a balanced Wheatstone bridge.

Derivation of Balanced Condition

Let the total current supplied by the cell be $I$. At junction $a$, the current splits into $I_1$ (flowing through branch $abd$) and $(I - I_1)$ (flowing through branch $acd$). The current passing through the galvanometer is denoted as $I_g$.

Applying Kirchhoff's Second Law (KVL) to closed loop $ABDA$:

$$I_1 P + I_g G - (I - I_1) R = 0 \quad \text{--- (i)}$$

Applying Kirchhoff's Second Law (KVL) to closed loop $BCDB$:

$$(I_1 - I_g) Q - (I - I_1 + I_g) S - I_g G = 0 \quad \text{--- (ii)}$$

Condition for Balance

When the bridge is balanced, the galvanometer shows no deflection, meaning the galvanometer current is zero ($I_g = 0$). Substituting $I_g = 0$ into equations (i) and (ii):

From equation (i):

$$I_1 P - (I - I_1) R = 0$$

$$I_1 P = (I - I_1) R \quad \text{--- (iii)}$$

From equation (ii):

$$I_1 Q - (I - I_1) S = 0$$

$$I_1 Q = (I - I_1) S \quad \text{--- (iv)}$$

Dividing equation (iii) by equation (iv):

$$\frac{I_1 P}{I_1 Q} = \frac{(I - I_1) R}{(I - I_1) S}$$

$$\frac{P}{Q} = \frac{R}{S}$$

Conclusion: This expression gives the condition for a balanced Wheatstone bridge. If the values of three resistors are known, the magnitude of the fourth unknown resistor can be precisely determined using this relation.

Fig: Wheatstone bridge circuit diagram showing resistor arms, galvanometer, and battery connections.

No comments:

Post a Comment