Physics, asked by raniso7779, 4 months ago

plz answer it fast ​

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Answered by Anonymous
7

 \large \underline \bold{Given}:-

\: \: \: \: \: \: \: \: \sf{V = 16 \: volt}

\: \: \: \: \: \: \: \: \sf{R1 = 2}Ω

\: \: \: \: \: \: \: \: \sf{R2 = 5}Ω

\: \: \: \: \: \: \: \: \sf{R3 = 3}Ω

 \large \underline \bold{To \: Find}:-

\sf{what's \: the \: potential \: difference \: across \: R1 \: , \: R2 \: and \: R3 \:. ?}

 \large \underline \bold{Usable \: Formula}:-

\sf{Series \: Combination \: of \: R1 \: , \: R2 \: , \: R3......Rn -}

\sf{then \: , }

\: \: \: \:  \small \bold{R = R1 + R2 + R3 + .....+ Rn}

\: \: \: \: \: \: \: \: \: \: \: \large\boxed{\sf\red{V = IR}}

 \large \underline \bold{Solution}:-

\sf{here \: ,}

R1 , R2 and R3 are in series combination .

\sf{So \: ,}

\sf{equivalent \: resistance-}

\: \: \: \: \sf{R = (2 + 5 + 3)}Ω

\: \: \: \:  \small \bold{R = 10}Ω

\sf{then \: -}

\: \: \: \: \: \: \sf{I = \dfrac{V}{R} = \dfrac{16}{10}}

\: \: \: \: \: \:  \small \bold{I = 1.6 \: A}

\sf{Now \: ,}

 \small \bold{potential \: difference \: across \: R1 -}

\: \: \sf{V1 = IR1 = 1.6\times 2 = 3.2 \: volt}

 \small \bold{potential \: difference \: across \: R2 -}

\: \: \sf{V2 = IR2 = 1.6\times 5 = 8 \: volt}

 \small \bold{potential \: difference \: across \: R3 -}

\: \: \sf{V3 = IR3 = 1.6\times 3 = 4.8 \: volt}

Answered by shobhanb701
0

Answer:

hi army.. saranghe, borahe sister

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