Question

show how would you connect 3 resistors of 6 oh m resistance so that the combination has a resistance of 9 ohm​

Answer 1

Answer:

HEYY!!!

Explanation:

connect two resistors(R1,R2) in parallel and then connect Rp with R3 in series

$\frac{1}{Rp} =\frac{1}{R1}+\frac{1}{R2} \\ \frac{1}{Rp} =\frac{1}{6}+\frac{1}{6} \\ \frac{1}{Rp} = \frac{2}{6} = \frac{1}{3}\\\\Rp=3ohm\\Rs=Rp+R3=3+6=9 ohm$

Answer 2

Explanation:

Let R1 , R2 , R3 be the resistors with resistance of 6 ohm respectively.

Let R1 || R2

Therefore resultant resistance -

1 / R = 1 / R1 + 1 / R2

1 / R = ( R2 + R1 ) / ( R1 x R2 )

R = ( R1 x R2 ) / ( R1 + R2 )

= ( 6 x 6 ) / ( 6 + 6 )

= 36 / 12

= 3

Let R3 be in series with the resultant of resistor R1 and R2

Therefore ,

R' = R + R3

= 3 + 6

= 9

Hence , to get a combination having resistance as 9 ohm , two resistances should be in parallel and the resultant of those resistances should be in series with the third resistance.

hope it helps you