Key relationship
P=VI=I²R=V²/R; Qheat=Pt.
Physics · Class 12
Electrical power and Joule heating
The circuit converts electrical energy into heat. With fixed resistance, doubling voltage doubles current and quadruples power.
Open the interactive experiment →Electrical power and Joule heating
மின்சக்தி P=VI. மின்தடை நிலையாக இருந்தால் மின்னழுத்தம் இரட்டிப்பானால் வெப்ப சக்தி நான்கு மடங்கு.
Key notes, relationships and model limits
Remember
The circuit converts electrical energy into heat. With fixed resistance, doubling voltage doubles current and quadruples power.
Common trap
At fixed current, P grows with R; at fixed voltage, P decreases with R. State the constraint before comparing heating.
Model limits
Constant resistances, all power becomes heat. Energy readout assumes exactly 10 s.
What you will learn
Use the controls and plotted relation to explain electrical power and joule heating.
Try the experiment
Try both scenarios, change one control, and predict the result before checking the graph and readouts.
Think of it this way
Change one input while keeping the others fixed, as you would in a controlled experiment.
Worked example
A 10 Ω resistor carrying 2 A dissipates 40 W and releases 400 J in 10 s.
Why this matters
The model connects electrical power and joule heating with an observable reading. Check the equation and assumptions before applying it.
Before you start
SI units · algebra · graph reading
Check your understanding
At fixed resistance, doubling current changes heating power by?
- A factor of two.
- A factor of four.
- A factor of one half.
Read the explanation
A factor of four.
Course connection: Class XII Chapter 3. Consult the roadmap for CBSE, NEET, JEE Main and Advanced scope.