An electric bulb is marked with 220 V, 50 W. Find the resistance of the filamentof thebulb. If the bulb is used 5 hours daily, find the energy in kilowatt-hour consumed bythe bulb in one month (30 days
Question
An electric bulb is marked with 220 V, 50 W. Find the resistance of the filamentof thebulb. If the bulb is used 5 hours daily, find the energy in kilowatt-hour consumed bythe bulb in one month (30 days
Solution
Sure, let's break this down into two parts.
Part 1: Finding the resistance of the filament of the bulb.
The power P of an electrical device (like a bulb) is given by the formula P = V^2 / R, where V is the voltage and R is the resistance. We can rearrange this formula to solve for R: R = V^2 / P.
Given that V = 220 V and P = 50 W, we can substitute these values into the formula:
R = (220 V)^2 / 50 W = 48400 V^2 / W = 968 ohms.
So, the resistance of the filament of the bulb is 968 ohms.
Part 2: Finding the energy consumed by the bulb in one month.
The energy E consumed by an electrical device is given by the formula E = P * t, where P is the power and t is the time.
Given that P = 50 W = 0.05 kW (since 1 W = 0.001 kW) and t = 5 hours/day * 30 days = 150 hours, we can substitute these values into the formula:
E = 0.05 kW * 150 hours = 7.5 kilowatt-hours.
So, the bulb consumes 7.5 kilowatt-hours of energy in one month.
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