Q.(a) Derive the condition of balance for Wheatstone bridge.

Wheatstone Bridge Balance and Meter Bridge

1. Balance Condition A Wheatstone bridge has four resistances $P$, $Q$, $R$, $S$ arranged in a quadrilateral with a galvanometer connecting opposite junctions and a battery across the other pair. At balance, no current flows through the galvanometer, so the two junctions it connects are at the same potential. This means the potential drops across adjacent arms are equal:

   $\frac{V_{AB}}{V_{AD}} = \frac{V_{CB}}{V_{CD}}$

   Since $V = IR$ and the currents through series arms are identical ($I_1$ through $P$ and $R$; $I_2$ through $Q$ and $S$), we get:

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

   Cross-multiplying gives the balance condition: $PS = QR$.

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

2. Meter Bridge Circuit …

Part (a): Deriving the Balance Condition

The Wheatstone bridge exploits a beautiful symmetry. When four resistors are arranged in a diamond configuration with a galvanometer connecting the middle nodes, we can adjust them until no current flows through the galvanometer. At that moment, the bridge is "balanced," and the resistances obey a simple ratio.

The Circuit Setup

Consider four resistances $P$, $Q$, $R$, and $S$ arranged as:

• $P$ and $Q$ in series along the top path (from point $A$ to $C$)
• $R$ and $S$ in series along the bottom path (from point $A$ to $C$)
• A galvanometer $G$ connecting the junction between $P$ and $Q$ (call it $B$) to the junction between $R$ and $S$ (call it $D$)
• A battery of emf $\mathcal{E}$ connected between $A$ and $C$

Why Balance Occurs

The galvanometer reads zero when points $B$ and $D$ are at the same potential. If they're at the same potential, there's no reason for current to flow between them.

Step-by-Step Derivation

1. Apply Kirchhoff's laws when the galvanometer current $I_g = 0$.

   Let current $I_1$ flow through $P$ and $Q$, and current $I_2$ flow through $R$ and $S$. Since no current flows through the galvanometer at balance, the same current that enters $P$ exits through $Q$, and likewise for $R$ and $S$.

2. Write the potential difference from $A$ to $B$ along the upper arm:

3. Write the potential difference from $A$ to $D$ along the lower arm:

4. Since $V_B = V_D$ at balance (no galvanometer current):

5. Now consider the potential from $B$ to $C$ and from $D$ to $C$:

6. Again, since $V_B = V_D$:

7. Divide equation (i) by equation (ii):

The currents cancel:

This is the balance condition for a Wheatstone bridge. Cross-multiplying gives the equivalent form $PS = QR$.

Part (b): Meter Bridge Circuit and Its Connection to Wheatstone Bridge

The meter bridge is a practical realization of the Wheatstone bridge using a uniform resistance wire.

Circuit Description

• A uniform resistance wire AC of length 100 cm (typically made of constantan or manganin) is stretched along a meter scale.
• Two gaps in the metallic strip below the wire: one contains the known resistance $R$, the other contains the unknown resistance $S$.
• These two resistances are connected in series between points $A$ and $C$ through the gaps.
• A galvanometer $G$ with one terminal connected to point $B$ (the junction between $R$ and $S$).
• The other terminal of the galvanometer is connected to a jockey (a sliding contact) that can touch the wire at any point $D$.
• A battery is connected across $A$ and $C$ with a key $K$.

How It Implements Wheatstone Bridge

The uniform wire AC acts as two resistances in series: …