Current & Power Dissipation in Two Bulbs in Series

[chawki]

Homework Statement

An ideal battery (internal resistance is zero) with E= 12,0 V is connected in series to two bulbs with resistances R1 = 2.0 Ω and R2 = 4.0 Ω.

Homework Equations

What is the current in the circuit and the power dissipation in each bulb

The Attempt at a Solution

R_eq=R₁+R₂
I=I₁=I₂=E/R_eq=12/6=2Amps.

P₁=R₁*I²=2*2²=8Watt.
P₂=R₂*I²=4*2²=16Watt.

 

[tiny-tim]

hi chawki!

An ideal battery (internal resistance is zero) with E= 12,0 V is connected in series to two bulbs with resistances R1 = 2.0 Ω and R2 = 4.0 Ω.
…
R_eq=R₁+R₂
I=I₁=I₂=E/R_eq=12/6=2Amps.

P₁=R₁*I²=2*2²=8Watt.
P₂=R₂*I²=4*2²=16Watt.

looks good!

(except it should be 2.0, 8.0, 16.0 )

 

[chawki]

ok thank you :) i will remember that.