Question

R₁ = 4 R₂ = 3 R₃ = 6 are connected in parallel the effective resistance (1/RP) is ....

Answer 1

S O L U T I O N :

Given,

• Three resistors R₁ = 4, R₂ = 3, R₃ = 6 are connected in parallel.

To Find,

• Effective resistance.

Explanation,

We know that,

1/Rp = 1/R₁ + 1/R₂ + 1/R₃

[ Put the values ]

=> 1/Rp = 1/4 + 1/3 + 1/6

=> 1/Rp = 3 + 4 + 2 / 12

=> 1/Rp = 9/12

=> 1/Rp = 0.75 ohm

Therefore,

The effective resistance is 0.75 ohm.

For information,

• Rs = R₁ + R₂ + R₃,.....
• V = IR
• H = I²Rt
• I = Q/t
• V = W/Q

Answer 2

Question ::-

R₁ = 4 R₂ = 3 R₃ = 6 are connected in parallel the effective resistance (1/RP) is ?

Given ::-

• $R_{1}$ = 4

• $R_{2}$ = 3

• $R_{3}$ = 6

• They are connected in parallel.

To find ::-

• Find the effective resistance ?

Formula used ::-

$⟹$ $\dfrac{1}{RP} = \dfrac{1}{ R_{1}} + \dfrac{1}{ R_{2}} + \dfrac{1}{ R_{3}}$

$⟹$ $\dfrac{1}{RP} = \dfrac{1}{ 4} + \dfrac{1}{3} + \dfrac{1}{ 6 }$

$⟹$ $\dfrac{1}{RP} = 3 +4+ \dfrac{2}{ 12 }$

$⟹$ $\dfrac{1}{RP} = \dfrac{9}{ 12 }$

$⟹$ $\dfrac{1}{RP} = 0.75 \: ohm$

______________

Hence , effective resistance = 0.75 ohm

let's explore more ::-

• R = $\dfrac{V}{I}$

• $R_{s}$ = $R_{1}$ + $R_{2}$ + $R_{3}$....

• $\dfrac{1}{RP} = \dfrac{1}{ R_{1}} + \dfrac{1}{ R_{2}} + \dfrac{1}{ R_{3}}$....