Kaveri conducted an experiment to study the energy efficiency of differentbulbs. She connected a bulb A having a resistance of 100 ohms to a 240 Vpower supply in a laboratory.(a) How much energy will be consumed by the bulb, if it is kept ON for 4hours each day for a week? Express your answer in kJ.(b) Kaveri connects another similar bulb B in series with bulb A and connectsthe combination to a 240 V supply. Will there be any change in the brightnesswith which bulb A glows now? Explain mathematically.

(a) First, we need to calculate the current (I) using Ohm's Law, which states that I = V/R, where V is the voltage and R is the resistance.

So, I = 240V / 100Ω = 2.4A

Next, we calculate the power (P) using the formula P = VI, where V is the voltage and I is the current.

So, P = 240V * 2.4A = 576W

Since power is the rate of energy consumption per unit time, we can calculate the energy consumed by the bulb in 4 hours by multiplying the power by the time (in seconds).

So, Energy = P * time = 576W * 4 hours * 3600 seconds/hour = 8294400J

Since 1 kJ = 1000 J, the energy consumed in kJ is 8294400J / 1000 = 8294.4 kJ

However, this is the energy consumed in one day. To find the energy consumed in a week, we multiply this value by 7 (the number of days in a week).

So, Energy consumed in a week = 8294.4 kJ/day * 7 days = 58060.8 kJ

(b) When another bulb B is connected in series with bulb A, the total resistance of the circuit increases. According to Ohm's Law (V = IR), for a constant voltage, if the resistance increases, the current must decrease.

Since the brightness of a bulb is directly proportional to the power consumed (P = VI), and both V and I have decreased, the power consumed by bulb A when bulb B is connected in series will be less than when it was connected alone. Therefore, bulb A will glow less brightly when bulb B is connected in series.