Current & Power Dissipation in Two Bulbs in Series

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The discussion revolves around calculating the current and power dissipation in two bulbs connected in series to an ideal battery with a voltage of 12.0 V. The total resistance is found to be 6.0 Ω, leading to a current of 2 Amps in the circuit. The power dissipation for the first bulb (2.0 Ω) is calculated to be 8 Watts, while the second bulb (4.0 Ω) dissipates 16 Watts. The calculations are confirmed to be correct, with minor adjustments noted for clarity. Overall, the key focus is on understanding series circuits and power calculations.
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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


Req=R1+R2
I=I1=I2=E/Req=12/6=2Amps.

P1=R1*I2=2*22=8Watt.
P2=R2*I2=4*22=16Watt.
 
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hi chawki! :wink:
chawki said:
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.
…
Req=R1+R2
I=I1=I2=E/Req=12/6=2Amps.

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

looks good! :biggrin:

(except it should be 2.0, 8.0, 16.0 :wink:)
 
ok thank you :) i will remember that.
 
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