JEE Main2013PhysicsCurrent ElectricityActual
The supply voltage to a room is 120 V . The resistance of the lead wires is 6 Ω . A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?
Options
- A13 .3   V
- B10 . 4   V
- Czero   V
- D2 .9   V
Correct answer
B. 10 . 4   V
Step-by-step solution
Resistance of bulb, R b = V 2 P b = 120 × 120 60 = 240   Ω Resistance of heater, R h = V 2 P h = 120 × 120 240 = 60   Ω Voltage across bulb before the heater is not connected, V 1 = V × R b R b + 6 = 120 × 240   V 246 = 117 . 07   V Now the bulb and heater are connected in parallel. Resistance of the combination is R n e t = 240 × 60 240 + 60 = 48   Ω Voltage across bulb after the heater is switched on, V 2 = V × R n e t R n e t + 6 = 120 ×