Question

Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on 120 V and has 39 remaining identical bulbs, what is then the operating voltage of each?

Short answer

There will be no light in the circuit and each bulb has a voltage of 3.1 Vacross it.

Step-by-step solution

Definition of voltage

Voltage: It's the difference between two points’ electric potentials.

Calculating voltage across each bulb

Current remains constant in a sequence. When a bulb burns out, current cannot flow to the following bulb because the burnt-out bulb acts as an open circuit. As a result, no other bulb can light up.

Voltage is distributed among the bulbs in sequence in a series. Because the bulbs have the same resistance, the voltage is distributed evenly among them.

The voltage across each bulb is given as,

$\mathrm{V}=\frac{{\mathrm{V}}_{\mathrm{battery}}}{\mathrm{n}}$

Substituting the given values, we get

$$\begin{gathered} \mathrm{V}=\frac{120\mathrm{V}}{40} \\ \mathrm{V}=3\mathrm{V} \end{gathered}$$

Calculating the normal operating voltage for each bulb

Current remains constant in a sequence. When a newer bulb burns out, the current can still flow to the next bulb because the burned-out bulb acts as a short circuit. As a result, other bulbs can light up. The total resistance of the circuit decreases as one of these bulbs burns out, resulting in a rise in current. The brightness of bulbs increases as the current increases.

Now when Number of bulbs in series:$\mathrm{n}=39$

The voltage of the battery:${\mathrm{V}}_{\mathrm{battery}}=120$volts

The voltage across each bulb $=\mathrm{V}$

$\mathrm{V}=\frac{{\mathrm{V}}_{\mathrm{battery}}}{\mathrm{n}}$

Inserting the values,

$$\begin{gathered} \mathrm{V}=\frac{120\mathrm{V}}{39} \\ \mathrm{V}=3.08\mathrm{V} \end{gathered}$$

Therefore, no bulb in the circuit will glow.

The voltage across each bulb comes out to be 3.1 V