Question

If five resistances, each of value 0.2 ohm, are connected in parallel, what will be the resultant

resistance?

Answer 1

Answer:

0.04 Ω

Explanation:

As per the provided information in the given question, we have :

• Five resistances, each of value 0.2Ω, are connected in parallel.

We are asked to calculate the resultant resistance.

As we know that, whenever the resistances are connected in parallel combination, then effective resistance is given by,

$\boxed{ \rm { \dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} +\dots \dfrac{1}{R_n} }}$

Here, we have :

• R₁ = 0.2 Ω
• R₂ = 0.2 Ω
• R₃ = 0.2 Ω
• R₄ = 0.2 Ω
• R₅ = 0.2 Ω

$\longmapsto \boxed{ \bf { \dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_4} + \dfrac{1}{R_5} }} \\ \\ \longmapsto \rm {\dfrac{1}{R_P} = \Bigg ( \dfrac{1}{0.2} + \dfrac{1}{0.2} + \dfrac{1}{0.2} + \dfrac{1}{0.2} + \dfrac{1}{0.2} \Bigg ) \; \Omega } \\ \\ \longmapsto \rm {\dfrac{1}{R_P} = \Bigg ( 5+ 5 +5+ 5+ 5 \Bigg ) \; \Omega } \\ \\ \longmapsto \rm {\dfrac{1}{R_P} = 25 \; \Omega } \\ \\ \longmapsto \rm {R_P = \dfrac{1}{25} \; \Omega } \\ \\ \longmapsto \underline{\boxed{ \bf {R_P = 0.04 \; \Omega }}} \; \bigstar$

∴ The effective resistance is 0.04 Ω.

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

More Information :

Whenever the resistances are connected in parallel combination, then effective resistance is given by,

$\bigstar \; \boxed{ \rm { \dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} +\dots \dfrac{1}{R_n} }}$

Whenever the resistances are connected in series combination, then effective resistance is given by,

$\bigstar \; \boxed{ \rm { R_S= R_1 + R_2 +\dots R_n }}$