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Answers
HI !
Method 1
Power = P = 100 W
Voltage = V = 220v
Resistance = R
P = V²/R
100 = 220 ×220/R
R = 220 × 220/100
= 484 Ω
==========================
Power = P = 60 W
Voltage = V = 220v
Resistance = R
P = V²/R
60 = 220 × 220/R
R = 220× 220/60
= 806.7 Ω
=========================
As the resistors are connected in parallel ,
total resistance = 1/R
1/R = 1/484 + 1/806.7
= 806.7 + 484/484×806.7
= 1290.7/390442.8
R = 390442.8/ 1290.7 = 302.5 ohms
Total resistance = 302.5 Ω
I = current
V = IR
220 = I × 302.5
I = 220/302.5
= 0.73 A
The current drawn is 0.73 A
OR
Method 2
Given :-
Resistance of 1st lamp, R₁ = V₁²/P₁ = (220)²/100
= 484 Ω
Resistance of 2nd lamp, R₂ = V₂²/P₂ = (220)²/60
= 806.6 Ω
Solution :-
Lamps are connected in parallel combination,
= R₁R₂/(R₁ + R₂)
= 484 × 806.6/(484 + 806.6)
= 390394.4/1290.6
= 302.5 Ω
Now, Current
I = V/Req
I = 220/(302.5)
I = 0.72 Ampere
Hence, The current drawn is 0.72 Ampere
Answer:
Power = P = 100 W
Voltage = V = 220v
Resistance = R
P = V²/R
100 = 220 ×220/R
R = 220 × 220/100
= 484 Ω
==========================
Power = P = 60 W
Voltage = V = 220v
Resistance = R
P = V²/R
60 = 220 × 220/R
R = 220× 220/60
= 806.7 Ω
=========================
As the resistors are connected in parallel ,
total resistance = 1/R
1/R = 1/484 + 1/806.7
= 806.7 + 484/484×806.7
= 1290.7/390442.8
R = 390442.8/ 1290.7 = 302.5 ohms
Total resistance = 302.5 Ω
I = current
V = IR
220 = I × 302.5
I = 220/302.5
= 0.73 A
The current drawn is 0.73 A
Hope it helps you
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