The Wheatstone bridge network shown in figure is balanced when δ = 0. For δ ≠ 0, the unbalanced voltage V_{0} is

This question was previously asked in

DRDO EE 2008 Official Paper

Option 1 : Eδ / 2

ST 1: Logical reasoning

5280

20 Questions
20 Marks
20 Mins

**Concept:**

**Wheat Stone Bridge:**

- A Wheatstone bridge is a special
**arrangement of 4 resistors**. - The device uses for the
**measurement of low-value resistance**by means of comparison of known resistance with the unknown resistance. - The Wheatstone bridge works on the
**principle of null deflection**, i.e. the ratio of their resistances are equal, and no current flows through the galvanometer. - The bridge is
**very reliable and gives an accurate result**.

Under balance condition,

\(\frac{P}{R} = \frac{Q}{S}\)

**Calculation:**

Given bridge circuit,

The bridge is balanced when δ = 0

∴ QR = PR

Q = P = x

**When δ ≠ 0:**

The bridge circuit can be drawn as,

\(\begin{array}{l} {I_1} = \frac{E}{{2x}}\\ {V_1} = {I_1}x = \frac{E}{2}\\ {I_2} = \frac{E}{{\left( {1 + \delta } \right)R + \left( {1 - \delta } \right)R}} = \frac{E}{{2R}}\\ {V_3} = {I_2}\left( {1 + \delta } \right)R = \frac{{E\left( {1 + \delta } \right)}}{2} \end{array}\)

From the circuit,

\({V_0} = {V_3} - {V_1} = \frac{{E\left( {1 + \delta } \right)}}{2} - \frac{E}{2}\)

\({V_0} = \frac{{E\delta }}{2}\)