Question

how can three resistance of 2,3 and 6 ohms be connected so as to give a total resistance of 1 ohm? show the working of your solution​

Answer 1

AnsWer :

Parallel Combination.

SolutioN :

Let's,

• R1 = 2Ω.
• R2 = 3Ω.
• R3 = 6Ω.

Now,

♢ Let's try to Connect all three resistance in Parallel Combination :

$\tt \dagger \: \: \: \: \: \dfrac{1}{R_p}= \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} ... \dfrac{1}{R_n}$

$\tt \dashrightarrow \: \: \dfrac{1}{R_p}= \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}$

$\tt \dashrightarrow \: \: \dfrac{1}{R_p}= \dfrac{3 + 2 + 1}{6}$

$\tt \dashrightarrow \: \: \dfrac{1}{R_p}= \dfrac{ \cancel6}{ \cancel6}$

$\tt \dashrightarrow \: \: \dfrac{1}{R_p}= 1\Omega.$

$\dashrightarrow \: \: \: \: \: \boxed{ \tt R_p= 1\Omega. }$

☛ Diagram :

----------------| 2Ω |-----------------

----------------| 3Ω |-----------------

----------------| 6Ω |-----------------

Therefore, the net resistance 1Ω which you get when we connect R1, R2 and R3 in Parallel Combination.

Answer 2

Given that, three resistors having resistance 2 ohm, 3 ohm and 6 ohm.

We have to connect the three resistors in such a way that their total resistance is 1 ohm.

For series combination:

Rs = R1 + R2 + R3

For parallel combination:

1/Rp = 1/R1 + 1/R2 + 1/R3

As the resultant resistance is minimum. So, connect them in parallel.

Take R1 = 2 ohm, R2 = 3 ohm and R3 = 6 ohm

Now, connect R1, R2 and R3 in parallel combination as shown:

------------| R1 = 2 ohm |-------------

------------| R2 = 3 ohm |-------------

------------| R3 = 6 ohm |-------------

1/Rp = 1/R1 + 1/R2 + 1/R3

Substitute value of R1, R2 and R3 in above formula,

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

LCM of 2, 3 and 6 is 6.

1/Rp = (3 + 2 + 1)/6

1/Rp = 6/6

1/Rp = 1/1

Rp = 1 ohm

Therefore, to get the total resistance of 1 ohm, we have to connect all the three (2, 3, 6 ohm) resistors in the parallel combination.