Question

Show how you would connect three resistors, each of resistance 6Ω, so that

the combination has a resistance of 9Ω.​

Answer 1

Given:-

• No. of Resistors = 3

• Resistance of Each Resistor ,R = 6 Ω

To Find:-

• Equivalent Resistance 9 Ω

Solution:-

We have to calculate the Equivalent resistance should be 9 ohms.

To get Equivalent Resistance 9 ohm the two Resistors must be connected in parallel and after that one Resistor should be connected in series then we get the Equivalent resistance 9 ohms . Let's Solve it

If two Resistors are connected in Parallel then Resistance in Parallel

• 1/Rp = 1/R1 + 1/R2

Substitute the value we get

→ 1/Rp = 1/6 + 1/6

→ 1/Rp = 2/6

→ Rp = 6/2

→ Rp = 3 ohm................(1)

Now, One Resistor is connected in series therefore, the Resistance in Series is

• Rs = R1 + ....

Substitute the value we get

→ Rs = 6 ohm ......................(2)

Additional equation (1) and (2) we get equivalent Resistance.

→ Req. = Rp + Rs

→ Req = 3 + 6

→ Req = 9 ohms .

Hence, required Answer is 9 ohms .

Answer 2

Answer:

Given :-

• No. of resistor = 3
• Resistance of each resistor = 6 ohm

To Find :-

Equivalent resistance

Solution :-

For calculating resistance of 9 resistor two of them must be in parallel.

$\sf \: \dfrac{1}{Rp} = \dfrac{1}{R1} + \dfrac{1}{R2}$

Putting values

$\sf \dfrac{1}{Rp} = \dfrac{1}{6} + \dfrac{1}{6}$

$\sf \dfrac {1}{Rp} = \dfrac{2}{6}$

$\sf \: Rp = \dfrac{6}{2}$

$\sf \: Rp = 3 ohm....(1)$

$\sf \: Rs = R1 +....$

$\sf \: Rs = 6 \: ohm(EQ 2)$

Now adding Equation 1 and 2

$\sf \: Req. = Rs + Rp$

$\sf \: Req. = 6 + 3$

$\sf \: Req = 9 \: ohm$