Question

1) when two resistors of resistance
R, and R2 are connected in llll the
het resistance is 3 ohm.
when connected in serius, its value is 16,
calculate the values of R, and R2.​

Answer 1

Answer:

$\mathfrak{\huge{\green{\underline{\underline {Answer:}}}}}$

 Resistance equivalent in parallel is given by R1 *R2 / (R1+R2)

Resistance equivalent in series is R1 + R2 

Given R1 + R2 = 16

and R eq in parallel as 3

sustitute the above values in the first equation of parallel equivalence 

we get R1*R2 = 3*16 = 48

The values of R1 and R2 are 12,4

By using basic calculations of (R1-R2)^2 = (R1 +R2)^2 - 4R1R2 

R1 - R2 = sqrt(256-192) = sqrt(64) = 8

solving R1 and R2 we get r1 =12 and r2 =4

Answer 2

$\large{\underline{\underline{\mathfrak{\green{\bf{\:QUESTION:-}}}}}}$

• 1) when two resistors of resistance
• R1, and R2 are connected in ll the
• het resistance is 3 ohm.
• when connected in series, its value is 16,
• calculate the values of R, and R2.

$\large{\underline{\underline{\mathfrak{\red{\bf{\:ANSWER:-}}}}}}$

$\large{\underline{\:GIVEN\:HERE:-}}$

• when two resistors of resistance
• R1, and R2 are connected in ll the
• het resistance is 3 ohm.

• when connected in series, its value is 16 .

$\bold{\underline{\:FIND\:HERE:-}}$

• Calculate the Value of resistance R1 and R2 .

$\large{\underline{\underline{\mathfrak{\red{\bf{\:EXPLANATION:-}}}}}}$

we know,

• When two resistance R1 and R2 are connected in parallel

So, Total resistance will be

$\bold{\boxed{\boxed{\:\frac{1}{R}\:=\:\frac{1}{R1}\:+\frac{1}{R2}}}}$

• when two resistance R1 and R2 are connected in series .

So, Total resistance will be

$\bold{\boxed{\boxed{\:R\:=\:R1\:+\:R2}}}$

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A/C to question ,

$\frac{1}{R1}\:+\frac{1}{R2}\:=\frac{1}{3}$

$\leadsto\:(\frac{R1+R2)}{R1R2})\:=\frac{1}{3}$

$\leadsto\:R1R2\:=\:3\times\:(R1+R2)$.....(1)

And,

$\:R1\:+\:R2\:=\:16$.....(1)

Keep value by (2) , in equation (2)

$\leadsto\:R1R2\:=\:3\times\:16$

$\bold{\boxed{\boxed{\:R1R2\:=\:48}}}$....(3)

But, we know,

• $\:(a-b)\:=\sqrt{(a+b)^2-4ab}$

So,

$\bold{\boxed{\boxed{(R1-R2)\:=\sqrt{(R1+R2)^2-4R1R2}}}}$

keep value by (1) and (3)

$\leadsto\:(R1-R2)\:=\sqrt{(16)^2-4\times\:48}$

$\leadsto\:(R1-R2)\:=\sqrt{256-\:192}$

$\leadsto\:(R1-R2)\:=\sqrt{64}$

$\bold{\boxed{\boxed{\:(R1-R2)\:=\:8}}}$....(4)

Now, we have

• R1 + R2 = 16

• R1 - R2 = 8

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Add this, we find

$\leadsto\:2R1\:=\:24$

$\leadsto\:R1\:=\cancel{\frac{24}{2}}$

$\bold{\boxed{\boxed{\:R1\:=\:12}}}$

keep value of R1 in (1),

$\leadsto\:(12+R2)\:=\:16$

$\leadsto\:R2\:=\:16-12$

$\bold{\boxed{\boxed{R2\:=\:4}}}$

$\large{\underline{\underline{\mathfrak{\bf{\:ANSWER:-}}}}}$

• Resistance of R1 = 12 ohm

• Resistance of R2 = 4 ohm

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$\large{\underline{\underline{\mathfrak{\bf{\:ANSWER\:VERIFICATION:-}}}}}$

Case(1):- when R1 and R2 in | | .

Total Resistance will be

$\implies\frac{R1R2}{(R1+R2)}\:=\:16$

keep value of R1 and R2,

$\implies\frac{12×4}{(12+4)}\:=\:3$

$\implies\frac{48}{16}\:=\:3$

$\implies\:3\:=\:3$

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Case (2) :- when R1 and R2 in series

Total resistance will be

$\implies\:(R1+R2)\:=\:16$

keep value of R1 and R2 ,

$\implies\:(12+4)\:=\:16$

$\implies\:(16)\:=\:16$

here, both case are satisfies ,

Hence, we can say that value of resistance are correct .

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