Question

$\Huge{\mathscr{\fcolorbox {navy}{orange}{\color {darkblue}{Question:-}}}}$

Prove that in a parallel connection the combine resistance is the sum of the reciprocals of all resistance.

$\boxed {\mathcal{\blue {\dfrac {1}{R}=\dfrac {1}{R_1}+\dfrac{1}{R_2}+\dfrac {1}{R_3}}}}$​​

Answer 1

Question :-

• Prove that in a parallel connection the combine resistance is the sum of the reciprocals of all resistance.

__________________

Solution :-

To Find :-

• Proof of :-

$\boxed {\bf{\red {\dfrac {1}{R}=\dfrac {1}{R_1}+\dfrac{1}{R_2}+\dfrac {1}{R_3}}}}$

Calculation :-

Since, the resistor are connected in parallel therefore the voltage will also be equal and same for all.

But, current will get divided,

⇒ I = I1 + I2 + I3.

Proceeding ahead,

$\implies (\sf\frac{V}{R}$) = ($\sf\frac{V}{R^1}$) + ($\sf\frac{V}{R^2}$) + ($\sf\frac{V}{R^3}$)

Using Ohm's Law,

⇒V ∝ I

⇒V = IR

⇒I = $\sf\frac{V}{R}$

Now we will take 'V' as common,

$\implies V (\sf\frac{1}{R}$) = V [($\sf\frac{1}{R1}$)+($\sf\frac{1}{R2}$)+($\sf\frac{1}{R3}$)]

$\implies \sf\frac{1}{R}$ = $\sf\frac{1}{R1}$+$\sf\frac{1}{R2}$+$\sf\frac{1}{R3}$

$\implies \therefore\boxed {\bf{\dfrac {1}{R}=\dfrac {1}{R_1}+\dfrac{1}{R_2}+\dfrac {1}{R_3}}}$

$\large \blue \dag \boxed{ \underline{ \underline{\tt \red {hence \: proved}}}} \blue\dag$

____________________

Note :-

Diagram in image attached above.

Answer 2

Answer:

Calculation :-

Since, the resistor are connected in parallel therefore the voltage will also be equal and same for all.

But, current will get divided,

⇒ I = I1 + I2 + I3.

Proceeding ahead,

\implies (\sf\frac{V}{R}⟹(

R

V

) = (\sf\frac{V}{R^1}

R

1

V

) + (\sf\frac{V}{R^2}

R

2

V

) + (\sf\frac{V}{R^3}

R

3

V

)

Using Ohm's Law,

⇒V ∝ I

⇒V = IR

⇒I = \sf\frac{V}{R}

R

V

Now we will take 'V' as common,

\implies V (\sf\frac{1}{R}⟹V(

R

1

) = V [(\sf\frac{1}{R1}

R1

1

)+(\sf\frac{1}{R2}

R2

1

)+(\sf\frac{1}{R3}

R3

1

)]

\implies \sf\frac{1}{R}⟹

R

1

= \sf\frac{1}{R1}

R1

1

+\sf\frac{1}{R2}

R2

1

+\sf\frac{1}{R3}

R3

1

\implies \therefore\boxed {\bf{\dfrac {1}{R}=\dfrac {1}{R_1}+\dfrac{1}{R_2}+\dfrac {1}{