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# the supply voltage to a room is 120V . The resistance of the lead wire is 6ohm . A 60W bulb is already switched on. What is the decrease of voltage across the bulb when a 240W heater is switched on in parallel to the bulb

Ashish Kumar
7 years ago
power=V^2/R V = 120
for bulb r1=120*120/60=240 ohms,for heater r2=120*120/240=60 ohms
Voltage before heater is connected =120*240/(6+240)=117.1 V
Voltage drop across bulb after heater is connected= 120*240/(6+240+60)=94.12  V
Decrease in voltage=117.1-94.12= 22.8V
465_96912_signature.GIF

Kalyan Awasthi
24 Points
5 years ago
How can 6 be added when in current division rule the total resistance of the closed loop is considered? (The above solution is incorrect)
Across bulb the drop is 117 Volts ( 120*240/(240+6) ) before the heater is ON.
Once the heater is ON SOME CURRENT WILL BE DRAWN BY HEATER ALSO.
Current drawn by the bulb will be (120/48+6)*(60/60+240)=0.44 A (Current division is used)
New drop across bulb = 0.44*240=106.66
Change in voltage across bulb = 117-106.66=10.34 V