Understanding Electric Circuits: What Happens When a Light Bulb Burns Out?

What happens to Bulbs A and C if Bulb B burns out?

Both Bulbs A and C will go out.

Bulb A will go out, but Bulb C will stay lit.

Bulb C will go out, but Bulb A will stay lit.

Both Bulbs A and C will stay lit.

Answer:

If Bulb B in a series circuit burns out, Bulbs A and C will also go out since the circuit will be incomplete. Holiday lights wired in series will all go out if one bulb burns out, but modern strings with shunts allow the rest to stay lit.

Explanation:

If Bulb B burns out in a series circuit, it interrupts the current flow in the circuit. Because the bulbs are connected in series, the current that flows through one bulb must flow through all others. Therefore, if Bulb B fails and behaves like an open switch, Bulbs A and C will also go out since there is no complete path for the electric current. This highlights the drawback of a series circuit where if one component fails, the entire circuit is affected.

In the case of holiday lights wired in series, if we assume that the old version utilizes bulbs that break the electrical connection when they burn out, this would mean that all the other bulbs will also go out. Conversely, with modern strings where the bulbs are designed to short circuit when they burn out thanks to a device called a shunt, the rest of the bulbs will stay lit because the current is able to bypass the burnt-out bulb.

For the holiday lights operating on 120 V with 40 identical bulbs, the normal operating voltage for each bulb would be 120 volts divided by 40 bulbs, which equals 3 volts per bulb. If a bulb in the newer string of holiday lights burns out and there are 39 remaining bulbs, each bulb would then operate at approximately 3.08 volts.