Final answer:
In a series circuit with multiple identical bulbs, the total voltage is divided evenly among the bulbs, so if a 30V battery is connected to three identical bulbs, each bulb receives 10V. If an old-style bulb in a series circuit burns out, the circuit opens and all bulbs go out; if a newer short-circuiting bulb burns out, the remaining bulbs receive slightly more voltage.
Step-by-step explanation:
If the voltage of the battery in the series circuit is 30V and there are multiple bulbs in the circuit, the voltage across each bulb will depend on how many bulbs are in the series. Assuming that the bulbs are identical and the internal resistance of the battery is negligible, the voltage will be evenly divided across each bulb. So, if we have, for instance, three bulbs, each bulb would receive 10V.
In a series circuit where some holiday lights are used, if an old version of a bulb that operates like an open switch when it burns out fails, the entire series is interrupted, and all bulbs would go out. Conversely, with newer versions where bulbs short circuit (operate like a closed switch) upon failure, the rest of the lights remain operational, but the voltage across each of the remaining bulbs would increase slightly, since the total resistance of the circuit would decrease.
For example, if a string operates on 120V and has 40 identical bulbs, each bulb would normally have an operating voltage of 3V (120V/40 bulbs). If one bulb burns out and it's a newer bulb that short circuits, now 39 bulbs must share the 120V, so each bulb would have approximately 3.08V (120V/39 bulbs).
Final answer:
In a series circuit with multiple identical bulbs, the total voltage is divided evenly among the bulbs, so if a 30V battery is connected to three identical bulbs, each bulb receives 10V. If an old-style bulb in a series circuit burns out, the circuit opens and all bulbs go out; if a newer short-circuiting bulb burns out, the remaining bulbs receive slightly more voltage.
Step-by-step explanation:
If the voltage of the battery in the series circuit is 30V and there are multiple bulbs in the circuit, the voltage across each bulb will depend on how many bulbs are in the series. Assuming that the bulbs are identical and the internal resistance of the battery is negligible, the voltage will be evenly divided across each bulb. So, if we have, for instance, three bulbs, each bulb would receive 10V.
In a series circuit where some holiday lights are used, if an old version of a bulb that operates like an open switch when it burns out fails, the entire series is interrupted, and all bulbs would go out. Conversely, with newer versions where bulbs short circuit (operate like a closed switch) upon failure, the rest of the lights remain operational, but the voltage across each of the remaining bulbs would increase slightly, since the total resistance of the circuit would decrease.
For example, if a string operates on 120V and has 40 identical bulbs, each bulb would normally have an operating voltage of 3V (120V/40 bulbs). If one bulb burns out and it's a newer bulb that short circuits, now 39 bulbs must share the 120V, so each bulb would have approximately 3.08V (120V/39 bulbs).