Question

theee resister of 6 ohm 3 ohm and 2 ohm are connected together so that total resistance is greater than 6 ohm but less than 8 ohm .draw a diagram to show this arrangement and calculate total resistance​

Answer 1

Answer:

Explanation:

if you put the 3 ohm and 2 om resistor in parallel then Rp would be

1/Rp= 1/3+1/2 = 5/6

Rp= 6/5 ohm = 1.2ohm

then connect them in series with the 6 ohm resistor

Rs = 1.2+6= 7.2 ohm which is greater than 6 and less than 8

Answer 2

$\sf \Large Solution!!$

Since total resistance is to be greater than 6 Ω, therefore 6 Ω resistor should be in series with the other resistors, the equivalent resistance of which is less than 2 Ω so that the total resistance is less than 8 Ω. Now, to have a resistance less than 2 Ω, remaining 3 Ω and 2 Ω resistors should be combined in parallel. The arrangement is shown in the attachment.

$\sf$

For resistors of 2 Ω and 3 Ω in parallel, the equivalent resistance $\sf R_{p}$ is given as

$\sf \dfrac{1}{R_{p}}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$

Therefore, $\sf R_{p}=\dfrac{6}{5}=1.2$Ω

The arrangement in the first attachment takes the the form as shown in the second attachment.

In the second attachment, the resistors of 6 Ω and 1.2 Ω are in series.

Therefore, total resistance of arrangement = 6 + 1.2 = 7.2 Ω (which is less than 8 Ω, but greater than 6 Ω).