19.7k views
2 votes
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
by
8.6k points

1 Answer

6 votes

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
by
7.5k points

Related questions

asked Jun 27, 2015 74.0k views
Virtuoz asked Jun 27, 2015
by Virtuoz
8.8k points
1 answer
3 votes
74.0k views
1 answer
2 votes
206k views