Science, asked by Brainly321117, 3 months ago

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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Answers

Answered by S4MAEL
175

\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}}


Ronak2141: Nice
ItzSantaclaus: yaa
rekhaprajapat3085: great
rekhaprajapat3085: I like it
rekhaprajapat3085: XD..
Anonymous: Fantastic bro ☺️
rekhaprajapat3085: yay
Itzdazzledsweetìe02: Perfect Explanation:Claps
Answered by Anonymous
213

Given that -

✠ There are two lamps that are connected in parallel to the electric main supply.

✠ One of the lamp rated as 100 W ; 220 V

Means,

R₁ = 220² / 100

✠ The another lamp rated 60 W ; 220 V

Means,

R₂ = 220²/60

To find -

✠ Current

Solution -

✠ Total current = 0.73 A

Using concept -

✰ Resistance (parallel connection).

✰ Ohm's law

Using formula -

✰ Resistance (parallel connection) = 1/R = 1/R₁ + 1/R₂ .... 1/Rn

✰ Ohm's law => V = IR

Where,

★ I denotes total current

★ R denotes resistance

Final solution -

~ Now,

➝ 1/R = 1/R₁ + 1/R₂ .... 1/Rn

➝ (1/220² / 100) + (1/220² / 60)

  • Cancelling the digits,

➝ R = 220² / 60

~ Using ohm's law

➝ V = IR

➝ V = 220²/106 I

➝ V = 48400/160 I

➝ 200 = 48400/160 I

➝ 200 = 4840/16 I

  • Cancelling the digits

➝ 200 = 605/2 I

➝ 200 = 302.5 I

➝ 200 / 302.5 = I

➝ 0.73 = I

➝ I = 0.73 A

Additional information -

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Kinetic \: energy \: is \: given \: by \: \dfrac{1}{2}mv^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Value \: of \: G \: is \: 6.673 \times 10^{-11}Nm^{2}kg^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Dimensional \: formula \: for \: universal \: gravitational \: constant \: is \: M^{-1} L^{3} T^{-2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto The \: unit \: of \: force \: constant \: k \: of \: a \: spring \: is \: \dfrac{N}{m}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Sir \: Cavendish \: was \: the \: first \: to \: gave \: value \: of \: G \: experimentally}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto The \: Young's \: modulus \: for \: perfect \: rigid \: body \: is \: infinite}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Density \: is \: the \: ratio \: of \: \dfrac{Volume}{Mass}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Maxwell \: is \: unit \: of \: magnetic \: flux}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: magnetic \: flux \: is \: Weber}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: surface \: tension \: is \: \dfrac{N}{m}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: mechanical \: power \: is \: Watt}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto 1 \: horsepower \: = \: approx \: 746 \: watts}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Momentum \: is \: measured \: as \: the \: product \: of \: Mass \: and \: velocity}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto \pi \: 'pi' \: is \: calculated \: by \: Aryabhatta}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto One \: J \: = \: 0.24 \: cal}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Number \: of \: SI \: units \: are \: 7}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Ampere \: is \: the \: unit \: of \: current \: electricity}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: Young's \: modulus \: of \: elasticity \: is \: Newton/m^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: pressure \: is \: Pascal}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Curie \: is \: the \: unit \: of \: radio \: activity}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Decibel \: is \: the \: unit \: of \: intensity \: of \: sound}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: electric \: charge \: is \: coulomb}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: resistance \: is \: ohm}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: acceleration \: is \: ms^{-2}}}}


ItzSantaclaus: Nice
NewBornTigerYT: Value of G = 6.673 * 10 ^ -11 Nm2/kg2
Anonymous: Thankies everyone ❤
Anonymous: And ya tq for saying I m correct it soon :)
rekhaprajapat3085: ok
Anonymous: Perfect ❤️
Anonymous: Shukriya ❤
Itzdazzledsweetìe02: Awesome as always
Anonymous: Shukriya ✌️(:
nishu200426: thnks
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