Question

A resistance R placed in parallel with another resistance of 40π ,their equivalent resistance is 24π .the value of R is..............​

Answer 1

Given:

• A resistance R is in parallel with another Resistance of 40 Ω.

• The equivalent resistance is 24 Ω .

To find :

• The value of R .

Solution :

• As we know that in parallel combination we find resistance by :

$\star \boxed{ \red{\bold{ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + .....}} }$

NOW ,

Let R1 = R and R2 = 40 Ω .

$\implies \: \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \\ \\ \implies \: \frac{1}{24} = \frac{1}{R} + \frac{1}{40} \\ \\ \implies \: \frac{1}{24} = \frac{40 +R }{40R} \\ \\ \implies \: 40R = 24(40 + R) \\ \\ \implies \: 40R = 960 + 24R \\ \\ \implies \: 40R - 24R = 960 \\ \\ \implies \: 16R \: = 960 \\ \\ \implies \: R = \cancel \frac{960}{16} \\ \\ \implies \: R = 60$

Hence ,the value of R is 60 Ω.

Answer 2

1/R = 1/R1+ 1/R2

1/24= 1/R + 1/40

1/24 = 40+R/ 40 R

R = 960/16 = 60