Question

Bulb b1 (100 w - 250 v) and bulb b2(100 w-200 v) are connected across 250 v. What is potential drop across b2?

Answer 1

Bulb B1: P = 100W

V = 250VR  

Resistance R1 = V2/P= (250×250)/100 = 625 ohm

Bulb B2: P = 100W

V = 200V

Resistance R2 = (200×200)/100 = 400 ohm

as both are in series combined Resistance R = 425+600 = 1025 ohm

Total current  I = total Voltage/ Total Resistance = 250/1025 = 10/41 A

potential difference Across B2 = R×I = 400× 10/41 = 4000/41 V

Answer 2

Answer:

Explanation:

Power = Voltage X Current

Voltage = Current X Resistance

P = V2/R

Bulb B1:

P= 100 W

V= 250 V

R1 = 625 ohms

Bulb B2:

P= 100W

V= 200V

R2= 400 ohms

Total Current = V'/ (R1+R2)

Here, V' = 250V (Main voltage supply)

R1+R2= 1025 ohms

I= 250/1025 = 0.244 A

Potential difference across B1 = I X R1 = 0.244 X 625 = 152.5 V

Potential difference across B2 = I X R2= 0.244 X 400 = 97.6 V