Heat Ventilation Air Conditioning Lectures

1. Analog devices (Wheatstone bridge)
ME-405 Measurement and instrumentation
Instructor : Dr Tabassum Yasmin
Department of Mechanical Engineering ,
University of engineering and technology Peshawar
Presented by : Muhammad Hammad
Registration no : 19pwmec4674
Section : E
Dated: 16-01-2023

2. Objectives:
• To understand the basic concepts of wheatstone bridge.
• To study the working principle of wheatstone bridge.
• To solve the problem related to the balancing of
wheatstone bridge.

3. Wheatstone bridge:
Wheatstone bridge, also known as the resistance bridge, calculates the unknown
resistance by balancing two legs of the bridge circuit. One leg includes the
component of unknown resistance.
The Wheatstone Bridge Circuit comprises two known resistors, one unknown resistor
and one variable resistor connected in the form of a bridge. This bridge is very
reliable as it gives accurate measurements.
Basic principle:
The Wheatstone bridge works on the principle of null deflection, i.e. the ratio of their
resistances is equal, and no current flows through the circuit. Under normal
conditions, the bridge is in an unbalanced condition where current flows through the
galvanometer. The bridge is said to be balanced when no current flows through the
galvanometer. This condition can be achieved by adjusting the known resistance and
variable resistance.

4. Problem statement (6.8):
A reactance bridge arrangement replaces the resistor in a Wheatstone bridge with a capacitor or
inductor. Such a reactance bridge is then excited by an AC voltage. Consider the bridge
arrangement shown in Figure below . Show that the balance equations for this bridge are given
by C2 ¼ C1R1/R2.
Required :
Show that the balance equations for this bridge are given by C2 = C1 R1/R2 is the requirement
for a balanced bridge.
Schematic diagram:

5. Solution:
With impedances (for an AC circuit)
𝑍1 = 𝑅1, 𝑍2 = 𝑅2 𝑍3 =
1
𝑗𝜔𝐶1
𝑍4 =
1
𝑗𝜔𝐶2
From equations (6.13) and (6.14)
Eo = I₁R₁+I₃R₃
Under these conditions, substituting Equations 6.9 to 6.11 into Equation 6.13 yields
𝐸𝜊 = 𝐸𝑖
𝑅1
𝑅1 + 𝑅2
+
𝑅3
𝑅3 + 𝑅4
So

7. Thank You !!!