FAD1022 L13 — Wheatstone Bridge and Voltage Divider

Week 5, Lecture #13 — This lecture covers two fundamental electrical measurement techniques: the voltage divider (potential divider) for creating variable voltages from a fixed source, and the Wheatstone bridge for precise resistance measurement.

13.1 Potential Divider/Voltage Divider

Definition

Voltage dividers (also known as potential dividers) are circuits that produce output voltage(s) as a fraction of their input voltage.

Main Purposes

  1. To yield a variable potential difference — adjustable output voltage
  2. To allow specific potential difference(s) to be selected — precise voltage selection
  3. To enable the splitting of potential difference of a power source between two or more components

Applications

1) Potentiometer

A passive electronic component that allows voltage to be adjusted by means of:

  • Sliding (linear potentiometer) — linear motion adjustment
  • Rotating (rotary potentiometer) — rotational adjustment

2) Level Shifters

Interfaces two circuits that use different operating voltages.

flowchart LR
    A[5V Circuit] -->|Voltage Divider| B[3.3V Circuit]
    
    subgraph "Level Shifter Example"
    C[5V Source] --- R1[1kΩ] --- D[Tapping Point<br/>3.3V Output]
    D --- R2[2kΩ] --- E[GND]
    end

13.2 Voltage Divider — Unloaded

Circuit Configuration

flowchart LR
    subgraph "Voltage Divider Circuit"
    A[V_in] --- B[R1] --- C[Tapping Point<br/>V_out] --- D[R2] --- E[GND]
    end

Derivation

Applying Kirchhoff's Loop Rule on Loop 1 (adcba):

$$-V_{in} + IR_2 + IR_1 = 0$$

$$V_{in} = IR_1 + IR_2 = I(R_1 + R_2) \quad \text{--- (1)}$$

$V_{out}$ corresponds to voltage drop across $R_2$:

$$V_{out} = V_{R_2} = IR_2 \quad \text{--- (2)}$$

From equation (1):

$$I = \frac{V_{in}}{R_1 + R_2} \quad \text{--- (3)}$$

Substituting (3) into (2):

$$V_{out} = \left(\frac{V_{in}}{R_1 + R_2}\right) \times R_2$$

Voltage Divider Formula

$$V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$$

Note: The term $\frac{R_2}{R_1 + R_2}$ is actually $R_p$ or $R_{eq}$ when considering parallel combinations.

Problem #13.1 — Unloaded Voltage Divider

Given:

  • $V_{in} = 100$ VDC
  • $R_1 = 25$ kΩ
  • $R_2 = 47$ kΩ

Find: Output voltage at point A with respect to ground

Solution:

$$V_{out} = V_{R_2} = V_{in} \times \frac{R_2}{R_1 + R_2}$$

$$V_{out} = 100 \times \frac{47000}{25000 + 47000} = 100 \times \frac{47000}{72000}$$

$$\boxed{V_{out} = 65.28 \text{ V}}$$

13.2 Voltage Divider — Loaded

Circuit Configuration

When a load resistance $R_L$ is connected across $R_2$:

flowchart LR
    subgraph "Loaded Voltage Divider"
    A[V_in] --- B[R1] --- C[V_out]
    C --- D[R2] --- E[GND]
    C --- F[R_L<br/>Load] --- E
    end

Analysis

$R_2$ and $R_L$ are in parallel. We can simplify the circuit:

$$R_p = \left(\frac{1}{R_2} + \frac{1}{R_L}\right)^{-1} = \frac{R_2 R_L}{R_2 + R_L}$$

The simplified circuit becomes $R_1$ in series with $R_p$.

Loaded Voltage Divider Formula

$$V_{out} = V_{in} \times \frac{R_p}{R_1 + R_p}$$

Or expanded:

$$V_{out} = V_{in} \times \frac{\frac{R_2 R_L}{R_2 + R_L}}{R_1 + \frac{R_2 R_L}{R_2 + R_L}}$$

Effect of Loading

Key Insight: When a load is connected:

  • $R_2$ and $R_L$ are in parallel
  • The total resistance at that part of the circuit becomes smaller ($R_p < R_2$)
  • As a result, voltage drop across $R_1$ becomes bigger compared to the unloaded case
  • The output voltage $V_{out}$ will be lower than the unloaded calculation would predict

Problem #13.2 — Loaded Voltage Divider

Question: What will happen to the voltages across resistors $R_1$ and $R_2$ when the load is connected to the divider circuit?

Solution:

What effect does adding the load have on the potential drop across $R_2$?

Since resistor $R_2$ and $R_L$ are parallel, it causes the total resistance at this part of the circuit to become smaller as compared to the value of $R_2$ alone, or $R_p < R_2$.

As a result, voltage drop across $R_1$ will become bigger (as compared to the case where no load is connected to $R_2$).

13.3 Wheatstone Bridge

Introduction

The Wheatstone bridge is a device used to measure resistances precisely by balancing potential drops in a circuit.

Circuit Components

A simple Wheatstone bridge consists of 4 resistors:

  • $R_1$ and $R_2$ — two known resistors
  • $R_3$ — a variable resistor
  • $R_x$ — the unknown resistor

The resistors are connected to a galvanometer and a voltage source.

flowchart TB
    subgraph "Wheatstone Bridge"
    A((a)) --- N[R1] --- B((b))
    A --- Rx[Rx<br/>Unknown] --- C((c))
    B --- R2[R2] --- D((d))
    C --- R3[R3<br/>Variable] --- D
    B --- G[Galvanometer<br/>G] --- C
    D --- S[Source] --- A
    end

Balance Condition

When the unknown resistor $R_x$ is placed in the circuit, the variable resistor $R_3$ is adjusted until the galvanometer reads zero.

Equilibrium is reached when galvanometer shows no deflection.

When there is no deflection on the galvanometer:

  • Points b and d are at the same potential
  • Zero potential difference across bd: $V_{bd} = 0$
  • Therefore, $I_{bd} = 0$

Derivation of Balance Condition

When the bridge is balanced:

Left branch: $I_{R_1} = I_{R_x} = I_1$

Right branch: $I_{R_2} = I_{R_3} = I_2$

Equal potential differences:

  • $V_{R_1} = V_{R_2}$ (between a-b and a-d)
  • $V_{R_x} = V_{R_3}$ (between b-c and d-c)

From Ohm's Law ($V = IR$):

$$I_1 R_1 = I_2 R_2 \quad \text{--- (1)}$$

$$I_1 R_x = I_2 R_3 \quad \text{--- (2)}$$

Dividing equation (1) by equation (2):

$$\frac{I_1 R_1}{I_1 R_x} = \frac{I_2 R_2}{I_2 R_3}$$

$$\frac{R_1}{R_x} = \frac{R_2}{R_3}$$

Wheatstone Bridge Formula

$$\boxed{R_x = \frac{R_1 \cdot R_3}{R_2}}$$

Or equivalently:

$$\boxed{\frac{R_1}{R_2} = \frac{R_x}{R_3}}$$

This can also be written as:

$$R_1 \cdot R_3 = R_2 \cdot R_x$$

Problem #13.3 — Slide Wire Bridge

Given:

  • A slide wire bridge is balanced
  • $R_1 = 15$ Ω
  • $R_2 = 5$ Ω
  • $R_3 = 9$ Ω

Find: Unknown resistance $R_x$

Solution:

$$R_x = \frac{R_1 \cdot R_3}{R_2} = \frac{15 \times 9}{5} = \frac{135}{5}$$

$$\boxed{R_x = 27 \text{ Ω}}$$

Problem #13.4 — General Wheatstone Bridge

Given:

  • Circuit with resistors M, N, P, and unknown X
  • At balance: M = 860 Ω, N = 14 Ω, P = 33.38 Ω

Part (a): Show that $X = \frac{MP}{N}$

Part (b): Determine the value of X

Solution (a):

When balanced:

  • $I_N = I_M = I_1$ and $I_P = I_X = I_2$
  • $V_N = V_P$ and $V_M = V_X$

From Ohm's Law:

$$I_1 N = I_2 P \quad \text{and} \quad I_1 M = I_2 X$$

Dividing:

$$\frac{I_1 N}{I_1 M} = \frac{I_2 P}{I_2 X} \Rightarrow \frac{N}{M} = \frac{P}{X}$$

Therefore:

$$\boxed{X = \frac{MP}{N}}$$

Solution (b):

$$X = \frac{(860)(33.38)}{14} = \frac{28706.8}{14}$$

$$\boxed{X = 2050.49 \text{ Ω}}$$

Summary of Key Formulas

Concept Formula
Unloaded Voltage Divider $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$
Parallel Resistance $R_p = \frac{R_2 R_L}{R_2 + R_L}$
Loaded Voltage Divider $V_{out} = V_{in} \times \frac{R_p}{R_1 + R_p}$
Wheatstone Bridge Balance $\frac{R_1}{R_2} = \frac{R_x}{R_3}$ or $R_x = \frac{R_1 \cdot R_3}{R_2}$

Related Concepts

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