Question

A resistor of 5 Ω is connected in series with a parallel combination of a number of resistor each of 5 Ω. If the total resistance of the combination is 6 Ω. How many resistors are in parallel.

Answer 1

Answer:

If the total resistance of the combination is 6Ω, then 5 resistors are in parallel combination.

Explanation:

Let the unknown number of resistors in parallel combination be :

➨ Y

In the question, we have :

➨ Resistor in series = 5Ω

➨ Resistance of resistor in parallel = 5Ω

➨ Resistance of total combination = 6Ω

We know that :

➨ Rs = R₁ + R₂ + R₃ . . . . Rₙ

➨ 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ . . . . 1/Rₙ

According to the question :

➨ 5Ω + (1/5Ω × Y) = 6Ω

➨ 5 + Y/5 = 6

➨ 5/1 + Y/5 = 6

➨ 25 + Y = 6 × 5

➨ 25 + Y = 30

➨ Y = 30 – 25

➨ Y = 5

Therefore, 5 resistors are in parallel combination.

Verification :

We know that :

➨ Total resistance = 6Ω

So, let's substitute and verify :

➨ 5Ω + (1/5Ω × Y) = 5Ω

➨ 5 + (1/5 × 5) = 6

➨ 5 + 1 = 6

➨ 6 = 6

LHS = RHS

Answer 2

Answer:

5

Explanation:

Given, first a 5 Ω resistor is connected with parallel combination of number of resistor of 5 Ω. It is also given that total resistance of the combination is 6 Ω.

We know that, resultant of parallel combination for identical resistors is

$\boxed{R_p = \dfrac{R}{x}}$

Here, Rp = effective resultant in parallel

R = Identical resistance

x = number of identical resistance

Here, R = 5

So,

$\implies R_p = \dfrac{5}{x}$

Now, this Rp is in series with a 5 Ω resistance

We know that, if R₁ and R₂ are in series, then Rs series effective resistance is,

$\boxed{R_s = R_1 + R_2}$

So, here, R₁ = 5 Ω and R₂ = 5/x and Total resistance(Here Rs) = 6 Ω

Thus,

$\implies 6 = 5 + \dfrac{5}{x}$

$\implies 1 = \dfrac{5}{x}$

$\therefore \underline{\boxed{x = 5}}$

Thus, number of resistors in parallel are 5.

[Diagram of circuit is in the attachment. Please refer it. All the resistors of 5 Ω]

Hope it helps!!