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Law Incorporation [45]
2 years ago
11

Q|C A lightbulb marked "75 W [at] 120 V " is screwed into a socket at one end of a long extension cord, in which each of the two

conductors has resistance 0.800ω . The other end of the extension cord is plugged into a 120-V outlet.(a) Explain why the actual power delivered to the lightbulb cannot be 75W in this situation.
Physics
1 answer:
VashaNatasha [74]2 years ago
3 0
  1. The actual power that is delivered to this lightbulb would be less than 75 Watts due to resistance of the conducting wires. Consequently, the resistance of these conducting wires would cause a voltage drop and as such reducing the power.
  2. The circuit diagram for the connection of this lightbulb is shown in the image attached below.
  3. The actual power delivered to this lightbulb is equal to 73.8 Watts.

<h3>How to calculate the power delivered to the lightbulb?</h3>

Mathematically, the power delivered to this lightbulb can be calculated as follows by using this formula:

Power, P = IV

Where:

  • I represents the current.
  • V represents the voltage.

Basically, the actual power that is delivered to this lightbulb would be less than 75 Watts due to resistance of the conducting wires. Consequently, the resistance of these conducting wires would cause a voltage drop and as such reducing the power.

Also, the circuit diagram for the connection of this lightbulb is shown in the image attached below.

<h3>How to determine the actual power delivered to this lightbulb?</h3>

First of all, we would determine the resistance in this electrical circuit:

P = IV = V²/R

R = V²/P

R = 120²/75

Resistance, R = 192 Ohms.

For the total equivalence resistance, we have:

Total equivalence resistance = 0.8 + 0.8 + 192

Total equivalence resistance = 193.6 Ohms.

Next, we would determine the current in this electrical circuit by applying Ohm's law:

V = IR

I = V/R

I = 120/193.6

Current, I = 0.62 Ampere.

Now, we can calculate the actual power delivered to this lightbulb:

P = I²R

P = 0.62² × 192

Actual power, P = 73.8 Watts.

Read more on actual power here: brainly.com/question/13896943

#SPJ4

<u>Complete Question:</u>

A lightbulb marked "75 W [at] 120 V" is screwed into a socket at one end of a long extension cord, in which each of the two conductors has resistance 0.800 V. The other end of the extension cord is plugged into a 120-V outlet.

(a) Explain why the actual power delivered to the lightbulb cannot be 75 W in this situation.

(b) Draw a circuit diagram.

(c) Find the actual power delivered to the lightbulb in this circuit?

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