Question

A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.

a) What is the current flowing through this light bulb?
b) What is the resistance of the light bulb?

Answer 1

bulb will burn out!

Step-by-step explanation:

Answer 2

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

`P = VI`

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

`I = (P)/(V) \\\\I = (100)/(220) \\\\I = 0.4545 \: A \\\\`

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

`V = IR`

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

`R = (V)/(I) \\\\R = (220)/(0.4545) \\\\R = 484 \: \Omega`

Therefore, the resistance of the bulb is 484 Ω