# Physics

In a series circuit there is 12V running with 3 bulbs with the following: 1Ω, 2Ω, and 3Ω.

Calculate the potential difference across each light bulb. Assume r = 0Ω.

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1. current in circuit=12/(1+2+3)=2a
potential across each bulb= I^2 R = 4R

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2. 6Ω total resistance , so 2A current

potential difference = current * resistance

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3. I = E/Rt = 12/(1+2+3) = 2A.
V1 = I * R1 = 2 * 1 = 2 Volts.
V2 = I * R2 = 2 * 2 = 4 Volts.
V3 = I * R3 = 2 * 3 = 6 Volts.

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