IIT JEE Physics (1978-2018: 41 Years) Topic-wise Complete Solutions

# Wheatstone bridge using Electric Bulbs!

## Introduction

A balanced Wheatstone bridge is very useful in analysing electrical circuits. The bridge is formed by five resistors of resistances $$R_1$$, $$R_2$$, $$R_3$$, $$R_4$$ and $$R_5$$. The circuit is balanced when $$R_1/R_2=R_3/R_4$$. In this case, the potential at points C and D are equal and no current flows through the resistor $$R_5$$. Generally, a galvanometer is connected in place of $$R_5$$.

## Apparatus

two bulbs of 100 watt, two bulbs of 60 watt, one bulb of 10 watt, five bulb holders and connecting wires.

## Procedure

In this demonstration, a Wheatstone bridge is constructed using light bulbs. Connect five bulb holders to make the circuit shown in the figure.

Plug two 100 watt bulbs in holder $$\mathrm{B_1}$$ and $$\mathrm{B_4}$$, two 60 watt bulbs in holder $$\mathrm{B_2}$$ and $$\mathrm{B_3}$$, and 10 watt bulb in holder $$\mathrm{B_5}$$. Connect the circuit to 220 V power supply. All the bulb starts glowing. Now, interchange the bulbs in $$\mathrm{B_1}$$ and $$\mathrm{B_3}$$. When plugged in, the four outer bulbs light while the central bulb doesn't. Try different combinations of the bulbs and holder and analyse the results. The same setup may be used to study series and parallel combinations of bulb. This is a good demo to eliminate misconception regarding $$P={V^2}/{R}$$ and/or $$P=i^2R$$.

Hazard: Be careful when working with 220 V AC supply. It may be fatal.

## Discussion

The resistance $$R$$ of the bulb is inversely proportional to its power $$P$$ i.e., $$R=V^2/P$$. Ask the students to calculate resistances of various bulbs and check Wheatstone bridge `balance condition' for various combinations.