Question

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2 bulbs rated 100 W - 220 V and 200 W - 220 V are connected in Parallel to a 220 V line. What is the total current drawn.​

Answer 1

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Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

GIVEN :

1st Bulb

P1 = 100 W, V1= 220 v

Now, we know

$P_{1}= \frac{ {V_{1} }^{2}}{R _{1}}$

$100 \: = \: \frac{ {220}^{2} }{R_{1}}$

$R _{1}\: = \: \frac{220 \times 220}{100}$

$R _{1} \: = \: 484 \: \Omega$

2nd Bulb

P2 = 200 W, V2 = 220 v

Now,

$P_{2}= \frac{ {V _{2}}^{2}}{R_{2}}$

$200 \: = \: \frac{ {220}^{2} }{ R _{2}}$

$R _{2} \: = \: \frac{ 220 \times \: 220}{200}$

$R _{2} \: = \: \frac{484}{2}$

$R _{2} \: = \: 242 \: \Omega$

Now, Total Resistance in Parallel

$\implies \: \frac{1}{R_{p}} = \: \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$= \: \frac{1}{484} + \frac{1}{242}$

$= \: \frac{1 + 2}{484}$

$= \: \frac{3}{484}$

$\therefore \: R_{p} \: = \: \frac{484}{3}$

Now, we know

$\: V = \: I \times R_{p}$

$\implies \: I = \frac{V}{R}$

$= \: \frac{220}{ \frac{484}{3} }$

$= \: \frac{220 \times 3}{484}$

$= \: \frac{30}{22}$

= 1.36 A

$\boxed{\sf{Hence, our \: required \: value \: is \: 1.36 \: A}}$