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

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  • 2 months ago
    Favourite 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.

  • 2 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.

  • 2 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.

  • oubaas
    Lv 7
    2 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

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  • 2 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

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