In a series circuit with a light and no capacitor to store charge, what happens when the switch is closed?

In a series circuit with a light and no capacitor to store charge, what happens when the switch is closed?

• A. It would be bright at first, then gradually dim and go out
• B. It remains bright indefinitely
• C. It never lights
• D. It flickers on and off

Answer: (A)

When the switch is closed in a simple series circuit, current starts flowing immediately and the lamp responds to that current. The filament is cold at the instant of switch-on, so its resistance is relatively low. That small moment lets a larger current pass, making the lamp appear brighter right away. As the filament heats up to its operating temperature, its resistance increases, which reduces the current and causes the brightness to fade. If the power source can’t sustain the current or is rapidly depleted, the light can eventually go out. Since there is no capacitor to store charge, there isn’t a charging transient to worry about—the change you see is due to the filament heating and the finite energy delivery from the source.

When the switch is closed in a simple series circuit, current starts flowing immediately and the lamp responds to that current. The filament is cold at the instant of switch-on, so its resistance is relatively low. That small moment lets a larger current pass, making the lamp appear brighter right away. As the filament heats up to its operating temperature, its resistance increases, which reduces the current and causes the brightness to fade. If the power source can’t sustain the current or is rapidly depleted, the light can eventually go out. Since there is no capacitor to store charge, there isn’t a charging transient to worry about—the change you see is due to the filament heating and the finite energy delivery from the source.