9 V with 3 Ω and 6 Ω: series vs parallel

A 9 V supply is connected to two resistors, R₁ = 3 Ω and R₂ = 6 Ω. What is the current in the circuit when they are connected in series, and when they are connected in parallel?

Step-by-step

1. 3 + 6 = 9
   Series resistances add: R = R₁ + R₂
2. 9 ÷ 9 = 1
   Ohm's law I = U / R
3. 1 ÷ (1/3 + 1/6) = 2
   Parallel resistances add reciprocally: 1/R = 1/R₁ + 1/R₂
4. 9 ÷ 2 = 4.5
   Ohm's law I = U / R (main current)

Why it works

In series the resistances add up, while in parallel the reciprocals add up and the total resistance is smaller than either resistor. The current then follows Ohm's law I = U / R.

Common mistakes

• Adding 3 Ω and 6 Ω when the resistors are in parallel.
• Expecting the parallel current to be smaller than the series current.

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