Potential difference (V) between two points in an electric circuit carrying current is defined as the work done (W) to move a unit positive charge (q) from one point to the other.
Mathematically: \[ V = \frac{W}{q} \]
The S.I. unit of potential difference is the Volt (V).
One volt is the potential difference between two points when 1 joule of work is done to move a charge of 1 coulomb: \[ 1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}} = 1 \text{ J/C} \]
Since:
Therefore: \[ \text{Volt (V)} = \frac{\text{Joule (J)}}{\text{Coulomb (C)}} \]
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).