Question

show how will you join three resistors each of 9 ohm so that equivalent resistance of combination is 13.5 ohm​

Answer 1

Answer:

When 2 resistors are connected in parallel combination and their equivalent resistance connected in series combination with 3rd resistor.

Explanation:

According to the given question,

R1, R2 and R3 = 9 ohm

And we are asked to find the combination by which we can have equivalent resistance of 13.5 ohm.

We know,

For parallel combination

• 1/R eq = 1/R1 + 1/R2

R eq is the equivalent resistance of R1 and R2 joined in parallel.

Putting the values we get,

• 1/R eq = 1/9 + 1/9

• 1/R eq = 2/9

• R eq = 9/2

• R eq = 4.5 ohm

Now this R eq is in series combination with R3, thus

For series combination

• R' eq = Req +R3

Where R' eq is the equivalent resistance of R eq and R3 joined in series.

Putting the values we get,

• R' eq = 4.5 + 9

• R' eq = 13.5 ohm

Therefore, R' eq is the final equivalent resistance.

Answer 2

Answer:-

Here,

$(R1, R2 ,R3 = 9 ohm )$Equivalent resistance what are of 13.5 ohm.

$\frac{1}{R} \: eq = \frac{1}{R1} + \frac{1}{R2}$

Equivalent resistance R eq

R1 and R2

Adding the values we have,

$\frac{1}{R} \: eq = \frac{1}{9} + \frac{1}{9}$

$\frac{1}{R} \: eq = \frac{2}{9}$

$R eq = \frac{9}{2}$

$R eq = 4.5 ohm$

$R eq = Req +R3$

R eq and R3 are connected

Again adding the values we have,

$Req = 4.5 + 9$

$Req = 13.5 ohm$

Hence, Req is equivalent resistance of final.