R1 ∥ R2
Give current a choice
Two branches share one supply. Watch their currents add up.
- R1 ∥ R2 · Parallel circuit
- Parallel circuit. Parallel branches connect to the same two nodes, so each receives 5 V. A node is a continuous electrical connection.
EXPERIMENT BENCH
Change the circuit settings, predict the readings, then check a challenge. These are learning models, with the assumptions explained in each lesson.
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Give current a choice
Ideal 5 V source with no current limit; real supplies have limits.
I(R1) = 5/R1 and I(R2) = 5/R2. Total current is their sum. A smaller branch resistance draws more current; adding a branch increases supply current.
Difficulty
Build · Give current a choice
Power on with both resistors at 1000 Ω. Observe 10 mA total.
Need a hint?
R1 = R2 = 1000 Ω
Explore · Give current a choice
Keep R1 = 1000 Ω; set R2 = 2000 Ω. Compare the two branch currents.
Need a hint?
R1 = 1000 Ω; R2 = 2000 Ω
Solve it · Give current a choice
Choose branch resistances for a total current of 15.00 mA.
Need a hint?
R1 = 500 Ω; R2 = 1000 Ω
Extra reading and circuit recipes
Give current a choice
Adding a branch reduces total resistance
Two 1000 Ω branches together behave like 500 Ω at the supply: each draws 5 mA, so the total is 10 mA. Equivalent resistance means the single resistance that would draw the same total current.
A real battery has limits
This experiment holds the supply at exactly 5 V. A real battery or supply cannot deliver unlimited current; its voltage can fall or protection can operate. Adding branches leaves each branch voltage unchanged only while the source maintains that voltage.
With R1 = 1000 Ω and R2 = 2000 Ω, which branch draws less current?
- R2
- R1
- They draw equal currents
Read more
Both branches have 5 V. R1 draws 5 mA and R2 draws 2.5 mA, because current equals voltage divided by resistance. Total current is 7.5 mA.
Change one branch, watch the other
Power on with both branches at 1000 Ω. Change only R2 to 500 Ω. Its current doubles from 5 to 10 mA, while R1 stays at 5 mA. Check the new total: 15 mA.