Question

Easy Question...!

Two lamps, one rated 100 W; 220V, and the other 60W; 220V, are connected in parallel to the electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220V.


_______

Just qualitied answer needed !

If u don't give proper as brainly rules then answer will deleted within some minutes' ​​

Answer 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

Answer 2

$\purple{\textsf{We know that,}}$

$\boxed{\pink{\mathsf{P = \frac{V^2}{R}}}}$

$\blue{\mathsf{\therefore R = \frac{V^2}{P}}}$

$\purple{\textsf{Resistance of the 1st lamp,}}$

$\implies \blue{\mathsf{R_1 = \frac{V^2}{P} = \frac{(220)^2}{100} = 484 \omega}}$

$\purple{\textsf{Resistance of the 2nd lamp,}}$

$\implies \blue{\mathsf{R_2 = \frac{(220)^2}{60} = \frac{2420}{3} \omega}}$

Since two lamps are connected in parallel, so its equivalent resistance is given by,

$\boxed{\pink{\mathsf{\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}}}}$

$\implies \blue{\mathsf{\frac{1}{R} = \frac{1}{484} + \frac{3}{2420} = \frac{8}{2480}}}$

$\implies \blue{\mathsf{R = \frac{2420}{8} \omega}}$

$\purple{\textsf{Current Drawn from the line,}}$

$\boxed{\pink{\mathsf{I = \frac{V}{R}}}}$

$\implies \blue{\mathsf{I = \frac{220 \times 8}{2420} = 0.73 A}}$

$\boxed{\red{\mathsf{Answer:- 0.73 A}}}$