Question

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

Answer 1

Answer:

bulb will burn out!

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,

$I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A$

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,

$R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega$

Therefore, the resistance of the bulb is 484 Ω