Question

14. If 2 resistors 412 and 812 are connected first in series and then in parallel. What is the ratio of their
effective resistances?
P.
A) 9:8
B) 3:2
C) 5:3
D) 9:2​

Answer 1

Appropriate Question :

If 2 resistors 4Ω and 8Ω are connected first in series and then in parallel. What is the ratio of their effective resistances?

Solution :

${\underline {\boxed {\Large \sf { Option \: D \: ( 9 \ratio 2 )} }}}$

Explication of steps :

According to the question,

$\to\boxed{ \sf{Ratio = \dfrac{Resistance_{(series)} }{Resistance_{(parallel)} }} }$

Effective resistance in series combination :

Effective resistance in series combination is given by,

• Rₛ = R₁ + R₂ + R₃ … Rₙ

In the given question,

● R₁ = 4Ω

● R₂ = 8Ω

● Rₛ = Effective resistance

So,

$\longmapsto$ Rₛ = R₁ + R₂

$\longmapsto$ Rₛ = 4Ω + 8Ω

$\longmapsto \boxed{\sf {R_s = 12 \Omega} }$

Effective resistance in parallel combination :

Effective resistance in parallel combination is given by,

• $\bf {\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots \dfrac{1}{R_n} }$

Here,

● R₁ = 4Ω

● R₂ = 8Ω

● Rₚ = Effective resistance

$\longrightarrow \sf {\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }$

$\longrightarrow \sf {\dfrac{1}{R_p} = \dfrac{1}{4} + \dfrac{1}{8} }$

$\longrightarrow \sf {\dfrac{1}{R_p} = \dfrac{2 + 1}{8} }$

$\longrightarrow \sf {\dfrac{1}{R_p} = \dfrac{3}{8} }$

$\longrightarrow \boxed{ \sf {R_p = \dfrac{8}{3} \Omega } }$

Calculating ratio of their effective resistance :

$\to\boxed{ \sf{Ratio = \dfrac{Resistance_{(series)} }{Resistance_{(parallel)} }} }$

$\to\sf{Ratio = 12 \div \dfrac{8}{3} }$

$\to \sf{Ratio = \cancel{ 12} \times \dfrac{3}{\cancel{8}} }$

• Divide 12 and 8 by 4.

$\to \sf{Ratio = 3 \times \dfrac{3}{2} }$

$\to \sf{Ratio = \dfrac{9}{2} }$

$\to \boxed{\pmb{\rm \red{Ratio = 9 \ratio 2}} }$

Therefore, the ratio of their effective resistances is 9:2. Option D is correct.

Answer 2

• 9:2 is our required ratio.

Explanation:

Correct Question:

If 2 resistors 4Ω and 8Ω are connected first in series and then in parallel, what is the ratio of their effective resistances?

Given:

• R₁ = 4Ω
• R₂ = 8Ω
• The resistors are first connected in series and then in parallel.

We need to:

• Find the ratio of their effective resistance.

Formulas used:

• When Resistors are connected in series, we use the formula: R₁+R₂
• When Resistors are connected in parallel, we use the formula: 1/R = 1/R₁+1/R₂

Concept: $\:$

Here, we need to use resistors in series formula at first. Then, we will use resistors in parallel formula and finally we will divide resistors in series by resistors in parallel to find the ratio of their effective resistance.

Solution:

When the resistors are connected in series, their effective resistance would be:

⟹ Effective Resistance = R₁+R₂

⟹ Effective Resistance = 4Ω+8Ω

⟹ Effective Resistance = 12Ω

Hence, effective resistance when resistors are connected in series will be 12Ω.

When the resistors are connected in parallel, their effective resistance would be:

⟹ 1/R = 1/R₁+1/R₂

⟹ 1/R = 1/4+1/8

On taking LCM, we get:

⟹ 1/R = (2+1)/8

⟹ 1/R = 3/8

On taking reciprocal, we get:

⟹ R = 8/3Ω

Hence, effective resistance when resistors are connected in series will be 8/3Ω.

According to the question, we need to find the ratio of their effective resistance. This means, we need to find:

⟹ Resistors in series : Resistors in Parallel

⟹ 12 : 8/3

⟹ 12×3 : 8

⟹ 36 : 8

⟹ 9 : 2

Hence, the required ratio is 9:2.