solve for x and y 11x+50y=133 50x+11y=172
Answers
Step-by-step explanation:
Question:
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Three identical bulbs are connected in parallel with battery the current drawn from the battery is 6 A. If one of the bulb gets fuse what will be the total current drawn from the battery?
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Answer:
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Total current drawn from the battery after one bulb gets fuse is 4 A.
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ㅤㅤㅤStep - By - Step Explanation
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\underline{\pmb{\bf{\bigstar\:\purple{Given\:that}\:.\:.\:.}}}★Giventhat...★Giventhat...
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Current in circuit = 6 A
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\underline{\pmb{\bf{\bigstar\:\purple{To\:Find}\:.\:.\:.}}}★ToFind...★ToFind...
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Total current drawn from battery if one bulb gets fuse?
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\underline{\pmb{\bf{\bigstar\:\purple{Required\:Solution}\:.\:.\:.}}}★RequiredSolution...★RequiredSolution...
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Let resistance of three bulbs be R, we know that they are connected in parallel. So, their equivalent resistance will be ::
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\bigstar\:\underline{\boxed{\pmb{\bf{\pink{\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}}}}}}}★Req1=R11+R21+R31Req1=R11+R21+R31
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\underline{\pmb{\bm{\bigstar\:Putting\:all\:values\:.\: . \:.}}}★Puttingallvalues...★Puttingallvalues...
\begin{gathered}\\ \qquad:\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} \\ \\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{1 + 1 + 1}{R} \\ \\ :\implies\:\sf \dfrac{1}{R_{eq}} = \dfrac{3}{R} \\ \\ :\implies\:\underline{\boxed{\pmb{\mathscr{\red{R_{eq} = \dfrac{R}{3}\:\text{\O}mega}}}}}\:\bigstar\end{gathered}:⟹Req1=R1+R1+R1:⟹Req1=R1+1+1:⟹Req1=R3:⟹Req=3RØmegaReq=3RØmega★
\begin{gathered}\\ \therefore\:{\underline{\sf{Hence,\:equivalent\:resistance\:is\:\pmb{\frac{R}{3}\:\text{\O}mega}}}}\end{gathered}∴