Two resistors- when connected in series have a total resistance of 25 - If they are connected in parallel- the total resistance reduces to 6 - - Find the values of resistances-

R_1+R_2=25Ω 1/R_1+1/R_2=1/R ⇒R_1+R_2/R_1R_2=1/6 ⇒ 6(R_1+R_2 )=R_1 R_2 ⇒ 6× 25=R_1 R_2 ⇒ R_1 R_2=150 ⇒ R_2=150/R_1...(1) Putting in (1) R_1+150/R_1=25 ⇒ R_1^2 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Equivalent Resistance of a System of Resistor

Two resistors...

Question

Two resistors, when connected in series have a total resistance of $25 Ω$ . If they are connected in parallel, the total resistance reduces to $6 Ω$ . Find the values of resistances.

Open in App

Solution

$R_{1} + R_{2} = 25 Ω$

$\frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{R}$
$\Rightarrow \frac{R_{1} + R_{2}}{R_{1} R_{2}} = \frac{1}{6}$
$\Rightarrow 6 (R_{1} + R_{2}) = R_{1} R_{2}$
$\Rightarrow 6 \times 25 = R_{1} R_{2}$
$\Rightarrow R_{1} R_{2} = 150$
$\Rightarrow R_{2} = \frac{150}{R_{1}} . . . (1)$
Putting in (1)
$R_{1} + \frac{150}{R_{1}} = 25$
$\Rightarrow R_{1}^{2} - 25 R_{1} + 150 = 0$
$\Rightarrow R_{1}^{2} - 15 R_{1} - 10 R_{1} + 150 = 0$
$\Rightarrow (R_{1} - 15) (R_{1} - 10) = 0$
$\Rightarrow R_{1} = 15 o r R_{1} = 10$
If $R_{1} 15 Ω, R_{2} = 10 Ω$ if $R_{1} = 10 Ω, R_{2} = 15 Ω$

Suggest Corrections

Similar questions

Two resistors having resistance $4 Ω$ and $6 Ω$ are connected in parallel. Find their equivalent resistance.

Three resistors of 4 Ω, 12 Ω , and 6 Ω are connected in parallel No. of 12 Ω resistors required to be connected in parallel to reduce the total resistance to half of its original is

Solve the following problems.

A. The resistance of a 1 m long nichrome wire is 6