100% 4 Rated
Attempts 127

60 W bulb on 110 V. Current drawn?

  1. A
    0.55 A
  2. B
    1.83 A
  3. C
    50 A
  4. D
    6 600 A

Topic Flashcards

Click to Flip
Question

What is the formula that relates electrical power (P), voltage (V), and current (I)? Rearrange it to solve for current.

Answer

P = I × V. Therefore, I = P / V.

Question

A 60 W light bulb is designed for a 110 V circuit. What is its operating current? If you mistakenly plug it into a 220 V outlet, what would happen to the current, assuming its resistance stays constant?

Answer

Operating current at 110 V: I = 60 W / 110 V ≈ 0.55 A. At 220 V, if resistance (R) is constant, current would double to about 1.1 A, and the bulb would draw about 240 W, causing it to burn out almost instantly.

Question

Using the power formula, calculate the resistance of the 60 W, 110 V light bulb filament during normal operation. (Hint: Use P = V²/R or find R from V and I).

Answer

R = V² / P = (110 V)² / 60 W ≈ 12,100 / 60 ≈ 202 Ω. Alternatively, I = 0.55 A, so R = V / I = 110 V / 0.55 A = 200 Ω (approximation differences are due to rounding).

Question

A device draws 0.55 A from a 110 V source. What is its power consumption in watts? How much energy in kilowatt-hours (kWh) does it use if left on for 10 hours?

Answer

Power = 0.55 A × 110 V = 60.5 W. Energy = 0.0605 kW × 10 h = 0.605 kWh

Question

An incandescent bulb is rated 60 W at 110 V. An LED bulb producing the same brightness is rated 9 W at 110 V. How much current does the LED bulb draw, and how does this illustrate improved efficiency?

Answer

I_LED = 9 W / 110 V ≈ 0.082 A. The LED provides similar light using ~85% less current (0.082 A vs. 0.55 A), meaning it converts much more electrical energy into light and less into heat.

Mini Quiz

1 / 3
For a given appliance, if the voltage is doubled, does the current drawn also always double?
Available Test Sets
Available FREE Test Sets