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.
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Answer 1

Step-by-step explanation:

$\underline\blue{\bold{ANSWER}}$:-

first lamp rated 100W , 220v (Given)

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

second lamp 60 W , 220v

$\large\ R_{2} = \frac{ {220}^{2} }{60}$(According to joule's law of heating - $\large\ p = \large\frac{ {V}^{2} }{R} )$

$\blue{\texttt{They\: are \: connected\: in\: parallel\: to\: the}}$

$\blue{\texttt{electric\: main \: supply\: so\: then,}}$

$\large\frac{1}{ R_{eq} } = \large\frac{1}{ R_{1}} + \large \frac{1}{ R_{2} }$

= $\large\frac{1}{ {220}^{2} /100 } + \large \frac{1}{ {220}^{2} / 60 }$

$R _{eq} = \large\ {220}^{2} /160$

from ohm's law,

$V = IR_{eq}$

$200 = \large\frac{ {220}^{2} }{160} I$

$\underline\blue{\bold{I = 0.73A}}$.

SO , $\underline\red{\bold{the \: current \: drawn \: by \: two \: bulbs \: from}}$

$\underline\red{\bold{the \: line \: is \:0.73A}}$

Answer 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