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An electric bulb is marked with 220V, 100W. Find the resistance of the filament of the bulb. If the bulb is used 5 hours daily, find the energy in kilo-watt hours consumed by the bulb in one month.

A. 484 ohms
B. 110 ohms
C. 44 ohms
D. 220 ohms

User Mihkel
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1 Answer

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Final answer:

The resistance of the 220V, 100W electric bulb filament is 484 ohms, and the energy consumed by using the bulb for 5 hours daily over a month will be 15 kilo-watt hours.

Step-by-step explanation:

To find the resistance of the filament in the bulb marked with 220V and 100W, we can use the formula for power:

P = V^2 / R, where P is the power in watts (W), V is the voltage in volts (V), and R is the resistance in ohms (Ω).

By rearranging the formula to solve for R, we get R = V^2 / P.

Substituting the given values:

R = 220V^2 / 100W = 48400 / 100 = 484 Ω.

The resistance of the filament is 484 ohms (Option A).

To calculate the energy consumption in kilo-watt hours (kWh) for 1 month, where the bulb is used 5 hours daily, we can use the formula:

Energy (kWh) = Power (kW) * Time (h).

The bulb's power in kW is 0.1 kW (since 100W = 0.1kW). The total hours used in a month (30 days) is 5 hours/day * 30 days = 150 hours.

Energy = 0.1 kW * 150 h = 15 kWh.

The bulb consumes 15 kWh of energy in one month.

User Eugenioy
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