Question

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 dia- gram.(c) Find the actual power delivered to the lightbulb in this circuit?

Answer 1

Answer:

Part c)

$P = 73.8 Watt$

Explanation:

Light bulb is marked as 75 W and 120 Volts

so it means that when we apply 120 volts across the bulb then it will consume the total power of 75 W

now when we connected the bulb with the help of conducting wires with some resistance across them then few of its voltage will fall across the wires

due to which the voltage across bulb is less than 120 Volts

So the power given by the bulb is less than 75 W

Part c)

Resistance of the bulb is given as

$R = \frac{V^2}{P}$

$R = \frac{120^2}{75}$

$R = 192 ohm$

now total resistance of the circuit

$R_{net} = 192 + 0.8 + 0.8$

now current in the circuit

$i = \frac{120}{193.6}$

$i = 0.62 A$

so actual power delivered to the bulb is given as

$P = i^2 R$

$P = 0.62^2 \times 192$

$P = 73.8 Watt$