The supply voltage to a room is 120V. The resistance of the lead wires is 6Ω. A 60W, 120V bulb is already switched on. What is the decrease of voltage (in Volts) across the bulb when a 240W, 120V heater is switched on in parallel to the bulb?

Resistance of bulb = (120 x 120) / 60 = 240Ω
Resistance of heater = (120 x 120) / 240 = 60Ω
When only the bulb is on:
Equivalent resistance = 6 + 240 = 246Ω
Current = 120 / 246 = 0.4878A
Voltage across bulb = 0.4878 x 240 = 117.07V
When both bulb and heater are on:
Equivalent resistance of bulb and heater in parallel = (240 x 60) / (240 + 60) = 48Ω
Equivalent resistance of the circuit = 6 + 48 = 54Ω
Total current = 120 / 54 = 2.22A
Current through bulb = (48/240) x 2.22 = 0.444A
Voltage across bulb = 0.444 x 240 = 106.56V
Decrease in voltage across the bulb = 117.07 - 106.56 = 10.51V
Therefore, the decrease in voltage across the bulb is approximately 10.4V