Light Bulbs in Series. A 60-ohm, 120-V light bulb and a 200-ohm,120-V light bulb

Question

Light Bulbs in Series. A 60-ohm, 120-V light bulb and a 200-ohm,120-V light bulb are connected in series across a 240-V line. Assume that the resistance of each bulb does not vary with current. (Note: This description of a light bulb gives the power it dissipates when connected to the stated potential difference: that is, a 25ohm, 120-V light bulb dissipates 25ohm when connected to a 120-V line.) (a) Find the current through the bulbs. (b) Find the power dissipated in each bulb. (c) One bulb burns out very quickly. Which one? Why?

Explanation / Answer

According to the standards used and also the line written in note that it dissipates 25 ohms which is incorrect, it is being taken as a print mistake and rather ohms is volts

Now, for 60 w bulb, 60 = 120^2/ r1 = 240 ohms

for 200w bulb, 200 = 120^2/r2 = 72 ohms

total circuit resistance = 240+72 = 312 ohms

(a) current through the bulbs = total current in circuit is i = v/r = 240/ 312 = 0.77 A

(b)power dissipated in 60w bulb = i^2/r1 = 0.77^2*240 = 142.3 watts ; power dissipated in 200w bulb = i^2/r2 = 0.77^2*72 = 42.68 watts

now, voltage across 60w bulb = i*r1 = 0.77*240 = 184.8 V and voltage across 200w bulb = i*r2 = 0.77*272 = 55.44 V

(c) since 60 w bulb is provided 184 w, it bursts quickly.

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