QUANTUM SHARMA · FREE SCIENCE & ENGINEERING LEARNING
Circuit Quest: learn electricity by playing. Start with a battery and a bulb, one clear step at a time.
For curious learners in India.
Begin with class 6–8 concepts.
No fees. No sign-up.
Choose a component, then check the result.
Try a choice and compare the measurement with your goal.
THE IDEA BEHIND THIS LEVEL
An open switch breaks the path. A closed switch connects it, allowing current around the loop.
Next: logic gates, then chip design.
Progress is saved in this browserChoose a component in the activity, read what changes, and check it against the mission. If the result misses the goal, try another choice. You can also learn the ideas here without using the controls.
LESSON 1 · A closed loop
A battery, a switch and a bulb make a simple circuit. A gap in the path stops current.
Your mission: Choose the switch position that completes the loop and lights the bulb.
An open switch breaks the path. A closed switch connects it, allowing current around the loop.
Close the switch to complete the path. Open it again to see why the bulb goes off.
LESSON 2 · Ohm’s law
The source supplies 6 V. For the resistor in this model, current I = voltage V ÷ resistance R.
Your mission: Choose a resistance that gives 0.30 A at 6 V.
Keep voltage fixed and compare the current with 10 Ω, 20 Ω and 30 Ω. Increasing resistance reduces current.
R = V ÷ I = 6 ÷ 0.30 = 20 Ω. At 10 Ω the current is 0.60 A; at 30 Ω it is 0.20 A.
LESSON 3 · Series and parallel
Two identical bulbs connect to an ideal 6 V source. Compare one series path with two parallel branches.
Your mission: Give each bulb the full 6 V.
Parallel branches share the same two connection points, so each has the full source voltage. Identical series bulbs divide it equally in this ideal model.
Choose parallel: each bulb has 6 V. In series each identical bulb has 3 V. This compares voltage, not real bulb brightness.
LESSON 4 · An LED and a resistor
In this model the LED has a fixed 2 V forward drop on a 5 V supply. The series resistor has the remaining 3 V.
Your mission: Choose a resistor for an LED current between 10 and 15 mA.
Use I = (5 − 2) ÷ R. A resistor limits current; an LED is not modeled as an ordinary fixed resistor.
220 Ω gives about 13.6 mA, within the target. 1,000 Ω gives 3.0 mA, below the target. With 0 Ω this model has no current limiting.
Build gates, solve eight challenges and create your own small circuits in Logic Lab.
Open Logic Lab →Explain current, voltage and resistance with worked examples and practice.
Learn more →Connect your circuit questions to the existing physics lessons.
Learn more →Read about matter, living things, energy and scientific evidence.
Learn more →