Derivation of the Balance Condition for a Wheatstone Bridge
A Wheatstone Bridge is used to measure unknown resistance by comparing it with known resistances. It consists of four resistances arranged in a diamond shape with a galvanometer connected between two opposite nodes. Let the resistances be:
- R1 and R2 in one branch - R3 and R4 in the other branch - A galvanometer connected between points B and D, and a battery connected across points A and C.
Step 1: Apply Kirchhoff’s Rules
Let the current through the galvanometer be IG. For the bridge to be balanced, the galvanometer shows zero deflection, so IG = 0. This implies that the potential at points B and D are equal:
\[ V_B = V_D \] Let the current through R1 be I1 and through R2 be I2. Similarly, current through R3 is I3 and through R4 is I4. Since no current flows through the galvanometer:
\[ I1 = I2 \quad \text{and} \quad I3 = I4 \] --- Step 2: Express Voltages Across Resistors
Potential difference across R1: \[ V_{AB} = I1 \, R1 \] Potential difference across R2: \[ V_{AD} = I2 \, R2 \] Potential difference across R3: \[ V_{AC} - V_{AB} = I3 \, R3 \] Potential difference across R4: \[ V_{AC} - V_{AD} = I4 \, R4 \] For a balanced bridge, the potentials at B and D are equal:
\[ V_B = V_D \implies I1 \, R1 = I3 \, R3 \quad \text{and} \quad I2 \, R2 = I4 \, R4 \] Since I1 = I2 and I3 = I4, the above simplifies to:
\[ \frac{R1}{R2} = \frac{R3}{R4} \] --- Step 3: Condition for Balance
\[ \boxed{\frac{R1}{R2} = \frac{R3}{R4}} \] This is the balance condition of a Wheatstone Bridge. When this condition is satisfied, the galvanometer shows zero current, indicating that the bridge is perfectly balanced.
--- Conclusion:
The Wheatstone Bridge is balanced when the ratio of resistances in one branch equals the ratio in the other branch. This allows precise measurement of an unknown resistance using the known resistances.
The terminal voltage of the battery, whose emf is\(10V\) and internal resistance\(1Ω\), when connected through an external resistance of \(4Ω\) as shown in the figure is:

A wire of length and resistance \(100\) is divided into 10 equal parts. The first \(5\) parts are connected in series while the next \(5\) parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
