Physics, asked by raveendarpalani, 1 day ago

28. Two resistors when connected in series, their effective resistance is 9 ohm. Their effective resistance become 2ohm when they are connected in parallel. Find the value of the resistors.​

Answers

Answered by MystícPhoeníx
54

Given: Two resistors when connected in series, their effective resistance is 9 ohm. Their effective resistance become 2ohm when they are connected in parallel.

Need to Find : Value of the Resistors

Let the value of resistors be x & y respectively

When these resistors connected in series the effective resistance is 9 ohm

x + y = 9 ------------------(i)

Now,

when these resistors connected in parallel their effective resistance is 2 ohm

1/x + 1/y = 1/2

x + y/xy = 1/2 -----------------(ii)

On substituting the equation (i) into equation (ii) we get

↠ 9/xy = 1/2

↠ 18 = xy

↠ 18/y = x ----------(iii)

Now, substituting equation (iii) in equation (i) we get

↠ 18/y + y = 9

↠ y² + 18 = 9y

↠ y² - 9y + 18 = 0

↠ y² - 6y - 3y + 18 = 0

↠ y ( y-6) + 3 ( y - 6 ) = 0

↠( y + 3) ( y-6) = 0

↠ y = -3 or y = 6

Since , resistance can't be negative .

So , y = 6

On substituting the value of y = 6 in equation (i) we get

↠ x + 6 = 9

↠ x = 9-6

↠ x = 3

  • Hence, the value of the resistors are 3 & 6 ohm respectively .

Answered by Anonymous
55

Information provided with us

  • Two resistors when connected in series, their effective resistance is 9 ohm.

  • Their effective resistance become 2 ohm when they are connected in parallel.

What we have to calculate

  • The required value of the resistor

Assumption

Consider the two resistance be

 \rm \: R_1

and

 \rm{R_2}

respectively

Now

 \rm \implies{Series \:  connection =R_s=R_1  + R_2 }

Or

\rm \implies{ 9=R_1  + R_2 }

\rm \implies{R_2  = 9 - R_1}

 \rm \implies{Parallel  \: connection  =  }

 \rm \mapsto \: R_p =  \dfrac{R_1 \: R_2}{R_1 \: R_2}

\rm \mapsto \: 2=  \dfrac{R_1 \: (9 - R_1)}{9}

\rm \mapsto \: R_{1}^{ \:  \: \: 2} +  9  \:  R_1 - 18 = 0

 \bf \implies \:  R_1 = 6  \:  \Omega

So

\bf \implies \: R_2 = 9 - 6 = 3 \:  \Omega

Therefore

 \rm \longrightarrow The \:  required \:  value \:  of  \: the \:  resistor \:  are \:  6   \: \Omega  \: and \: 3 \: \Omega

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