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

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,

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

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,

Therefore, the resistance of the bulb is 484 Ω
Answer:
Thats her fault.........................b
Explanation:
A closed circle means the number is included and an open circle means its not.
A single photon carries an energy equal to

where h is the Planck's constant and f is the frequency of the photon.
This means that the higher the frequency of the light, the higher the energy. Among the 5 different options mentioned by the problem, the light with highest frequency is ultraviolet, which has frequencies in the range [3-30] PHz, while visible light (red, blue, green) and infrared have lower frequency, so ultraviolet light has the highest energy per photon.