Bulb B1(100W- 250V) and bulb B2 …

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Bulb B1(100W- 250V) and bulb B2 (100W-200V) are connected across 250V in series.What is potential drop across B2 ?

Posted by Pravesh Kundu 8 years, 10 months ago

CBSE > Class 10 > Science

2 answers

Naveen Sharma 8 years, 10 months ago

Ans. 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

1Thank You

Shweta Gulati 8 years, 10 months ago

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

1Thank You

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Posted by Pandey Ji 5 months, 3 weeks ago

CBSE > Class 10 > Science