# Circuit question (physics)?

Two filament lamps a1 and a2 of nominal capacity a1=40W and a2=60W are designated for same voltage U0, connected into series electric circuit with voltage of U0. What is the power each lamp will use? Which one will shine brighter?

Sorry for my english

### 5 Answers

- micatkieLv 72 months agoFavourite answer
Resistance of a₁, R₁ = Uₒ²/P₁ = Uₒ²/40 Ω

Resistance of a₂, R₂ = Uₒ²/P₂ = Uₒ²/60 Ω

Equivalent resistance of the two lamps, R = (Uₒ²/40) + (Uₒ²/60) = 0.416Uₒ²

Current of the series current, I = Uₒ/R = Uₒ/(0.416Uₒ²) = 24/Uₒ A

In the series circuit:

Power of a₁ = I²R₁ = (24/Uₒ)² × (Uₒ/40) = 14.4 W

Power of a₂ = I²R₂ = (24/Uₒ)² × (Uₒ²/60) = 9.6 W

a₁ will shine brighter because of the higher power.

- oldschoolLv 72 months ago
P = V²/R so R = V²/P = Uo²/P

The current i = Uo/(Uo²/40 + Uo²/60) =

Uo/(60Uo²/2400 + 40Uo²/2400) = 2400Uo/Uo²(60+40)

= 24/Uo = i in the ckt with the two resistors in series with source Uo. Thus the power in the 40W

= 24²/Uo² *Uo²/40 = 24²/40 = 14.4W

In the 60W P = 24²/60 = 9.6W so the 40W bulb is brighter.

- PhilomelLv 72 months ago
Each lamp will have 1/2 design voltage.

The 40W a1 will dissipate ~10W.

The 60W a2 will dissipate ~15W.

The 60W bulb a2 will be brighter.

- oubaasLv 72 months ago
R1 = Uo^2/40

R2 = Uo^2/60

I = Uo/(R1+R2) = Uo*120/5Uo^2 = 24/Uo

P1 = R1*I^2 = Uo^2/40*24^2/Uo^2 = 14,4 watt

P2 = R2*I^2 = Uo^2/60*24^2/Uo^2 = 9,6 watt

total P = 24 watt

or more simply :

equivalent power in series is the parallel of the powers, i.e. 60 // 40 = 24 watt

P1 = P*a1/(a1+a2) = 24*60/100 = 14,4 watt

P2 = Pa2/(a1+a2) = 24*40/100 = 9,6 watt

a1 will shine more than a2

- What do you think of the answers? You can sign in to give your opinion on the answer.
- billrussell42Lv 72 months ago
filament lamps are very non-linear vs voltage, so this cannot be solved.

BUT, if I assume they are linear...

power = voltage²/resistance

for the first, 40 = U₀²/R₁

for the second, 60 = U₀²/R₂

R₁ = U₀²/40

R₂ = U₀²/60

note that R₂ < R₁

now put them in series. Use voltage divider rule.

U₁ = U₀R₁/(R₁+R₂)

U₂ = U₀R₂/(R₁+R₂)

power = U²/R

P₁ = (U₀R₁/(R₁+R₂))²/R₁

P₂ = (U₀R₂/(R₁+R₂))²/R₂

P₁ = R₁(U₀/(R₁+R₂))²

P₂ = R₂(U₀/(R₁+R₂))²

now compare the two.

(U₀/(R₁+R₂))² term is same, so power is proportional to resistance

since R₂ < R₁, that means P₂ < P₁ and A₁ is brightest