Question

plzz Solve it..... ​

Answer 1

$\large \underline \bold{Given}$:-

$\: \: \: \: \: \: \: \:$$\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}$

Answer 2

Answer:

R1 , R2 and R3 are in series combination .

So ,

equivalent resistance

$\: \: \: \:$$\sf{R = (2 + 5 + 3)}$Ω

$\: \: \: \:$$\small \bold{R = 10}$Ω

then : -

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

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

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}$