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

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
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