Question

1. the resistance of a tungsten bulb increases as it gets hotter. how does this help explain why many bulbs burn out when they are first turned on?

Explanation:

When a bulb is first turned on, its temperature is low, so resistance (R) is low. By Ohm's Law ($I = \frac{V}{R}$), with constant voltage (V) from the source, low R means high initial current (I). High current causes more power ($P = I^2R$) dissipation, stressing the bulb's filament, leading to burnout.

Answer:

When a bulb is first turned on, its temperature (and thus resistance) is low. By \( I = \frac{V}{R} \) (constant \( V \)), low \( R \) gives high initial current. High current (\( P = I^2R \)) causes excessive power, stressing the filament and leading to burnout.