A 60-watt lightbulb is powered by a 110-volt power source. What is the current being drawn?

A
0.55 amperes

B
1.83 amperes

C
50 amperes

D
6,600 amperes

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Question

Using the formula P = I × V, you calculated a current of ~0.55 A. Now, use Ohm's Law (V = I × R) to find the approximate resistance of the lightbulb's filament when it is operating.

Answer

R = V / I = 110 V / 0.55 A ≈ 200 Ω. This connects power formulas to Ohm's Law and gives a tangible property (resistance) of the bulb.

Question

If you plugged this 60-watt, 110-volt bulb into a 220-volt outlet by mistake, what would initially happen to the current drawn? (Assume the resistance stays roughly the same).

Answer

The current would be I = V / R = 220 V / 200 Ω = 1.1 A. This is double the intended current, which would cause the bulb to burn out almost instantly. This teaches the critical concept that devices are designed for specific voltages.

Question

A "60-watt" rating tells you the bulb's power consumption at its designed voltage. How much electrical energy (in Joules) does this bulb use if it is left on for 5 hours?

Answer

Energy = Power × Time. Convert time: 5 hours × 3600 s/hour = 18,000 s. Calculate: 60 W × 18,000 s = 1,080,000 J or 1.08 MJ. This links the common unit of "watts" to the fundamental unit of energy (joules) and practical usage.

Question

A modern 9-watt LED bulb produces the same light as this old 60-watt incandescent. If both are powered by 110 volts, how does the current drawn by the LED compare?

Answer

I_LED = P / V = 9 W / 110 V ≈ 0.082 A. The LED draws only about 15% of the current (0.082 A vs. 0.55 A) to produce the same light, illustrating the dramatic efficiency improvement of newer technology.

Question

In the formula P = I × V, if you increase the voltage while keeping the power constant (like in efficient devices), what must happen to the current?

Answer

The current must decrease. Since P is held constant, I = P / V. This is an inverse relationship. This explains why high-voltage power lines are used for transmission: for the same power, higher voltage means lower current, which reduces energy loss in the wires.

Mini Quiz

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Does the current always equal the power (in watts) divided by the voltage (in volts) for any electrical device?

A 60-watt lightbulb is powered by a 110-volt power source. What is the current being drawn?

A
0.55 amperes

B
1.83 amperes

C
50 amperes

D
6,600 amperes

The current drawn by the lightbulb is 0.55 amperes.

Topic Flashcards

Mini Quiz

Related Questions

A 60-watt lightbulb is powered by a 110-volt power source. What is the current being drawn?

A 0.55 amperes B 1.83 amperes C 50 amperes D 6,600 amperes

The current drawn by the lightbulb is 0.55 amperes.

Topic Flashcards

Mini Quiz

Related Questions

A
0.55 amperes

B
1.83 amperes

C
50 amperes

D
6,600 amperes