Question

Two resistors are connected in series gives an equivalent resistance of 10 Ω. When connected in parallel, gives 2.4 Ω. Then the individual resistance are:
(a) each of 5 Ω
(b) 6 Ω and 4 Ω
(c) 7 Ω and 4 Ω
(d) 8 Ω and 2 Ω

Answer 1

Answer

6Ω and 4Ω

Explanation:

Hello There

Since the resistors in series gives total resistance as 10Ω

R1+R2=10Ω

And in parallel it gives 2.4Ω

$\frac{R1*R2}{R1+R2}$=2.4Ω

Which implies

$\frac{R1*R2}{100}=2.4$Ω

R1*R2=24Ω

for R1+R2=10

and R1*R2=24

R1=6 and R2=4

Answer 2

Answer:

(b) 6 ohm and 4 ohm

Explanation:

let,

The resistors be R1 and R2 respectibely.

for series combination,

R = R1 + R2

R1 + R2 = 10 ________ ( 1 )

Now,

1/R = 1/R1 + 1/R2

1/2.4 = 1/R1 + 1/R2 _______ ( 2 )

By solving ( 1 ) & ( 2 ) :

R1 = 6 ohm

R2 = 4 ohm