Respuesta :
The problem describes the relationship of "bulb a" and "bulb b" to be in connected in series. When the switch is open then no current can flow, on the other hand, when it is closed, current will pass through.
When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.
Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.
So the light of "bulb a" will be dimmer.
When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.
Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.
So the light of "bulb a" will be dimmer.
There is a pattern of current flow. Based on the above scenario, Option b is the correct answer as bulb B will be brighter than before (Image attached).
What does this implies?
Based on the current flowing through bulb A is said to remains the same in both scenario. It is the current through bulb B that changes when the switch are said to be closed.
There is something that happens to the brightness of a bulb when the switch is said to be closed. Note that when the switch is closed, the light bulb often operates because of the current flows via the circuit. The bulb (B) tends to glows at its full brightness because it receives more volts.
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