Question

state how would you connect the 3 resistance 60hms so that the combination has a resistance of 90 hms​

Answer 1

Explanation:

ANSWER

We are given 3 resistors each of resistance 6 Ω. Equivalent resistance of resistors in series is more than the individual resistance of each resistor while equivalent resistance of resistance in parallel is less than the individual resistance of each resistor.

(i)

To get an equivalent resistance of 9 Ω, atleast one six ohm resistor should be connected in series with the other resistors.

Since, R

eq

=R

1

+R

2

9=6+R

2

R

2

=3 Ω

⟹R

2

<6 Ω

Connecting the remaining two resistors in parallel will yield an equivalent resistance of 3 Ω.

Hence, the first figure shows the configuration of resistors to get equivalent resistor of 9 Ω.

(ii)

To get an equivalent resistance of 9 Ω, no six ohm resistor should be connected in series with the other resistors.

Let one of the branches have only one 6 Ω resistor.

Since,

R

eq

1

=

R

1

1

+

R

2

1

9

1

=

6

1

+

R

2

1

R

2

=12 Ω