You have a string of holiday lights, consisting of 30 bulbs in series. The strin

Question

You have a string of holiday lights, consisting of 30 bulbs in series. The string of lights is plugged into a wall socket, which we can assume to supply a voltage of 120 V. There is also a fuse in series with the light bulbs. The fuse is designed to blow (this means the current cannot flow anymore) if the current is greater than or equal to 3.0 A; however, with 30 bulbs the current is only 2.30 A If a bulb in the string burns out, what happens is that a short circuit is formed that allows the current to bypass the burned-out bulb. The short circuit is a path of negligible resistance, in parallel with the burned-out bulb. This allows the remaining lights to stay on, but at the expense of increasing the current in the circuit, and increasing the chance that more bulbs will burn out. One of the purposes of the fuse is to prevent all the bulbs from burning out. (a) If one bulb burns out, so the string goes from having 30 bulbs in series to 29 bulbs in series, by how much will the current increase? Note that we're not asking for the new current. Instead, we're looking for the difference between the new current and the original current. (b) Note that if another bulb burns out, the current will increase by a different amount. Starting with the original 30 bulbs, how many bulbs on the string have to burn out before the fuse blows? bulbs

Explanation / Answer

I = V / Req

2.30 = 12 / (30 R)

R = 0.174 ohm

(A) I = 12 / (29 x 0.174) = 2.38 A

(B) 3 = 12 / (0.174n)

n = 23

so bulbs burn out = 7

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