Question

Question :

$\mathsf{ If \: equivalent \: resistances \: of \: R_1 \: and \: R_2 \: in \: series }$
$\sf{ \: and \: parallel \: be \: r_1 \: and \: r_2 \: respectively, \: then}$
$\sf{ \: \dfrac{R_1}{R_2} \: equals }$

Options :

$\sf{(a) \: \dfrac{r_1}{r_2} \qquad \qquad (b) \dfrac{r_1r_2}{r_1 +r_2 }}$
$\sf{(c) \dfrac{r_1 + \sqrt{ {r_1}^{2} - 4r_1r_2 } }{r_1 + \sqrt{{r_1}^2+4r_1r_2}} \qquad \qquad (d) \dfrac{ r_1 + \sqrt{{r_1}^2-4r_1r_2 }}{r_1 - \sqrt{{x_1}^{2} -4r_1r_2}}}$

Answer with proper explanation !

Be Brainly ^_^​

Answer 1

✿$\underline{\textbf{\textsf{ \purple{Solution}:- }}}$

$\sf{\\}$

$\underline{\underline{\textbf{ \red{In series}:- }}}$

$\sf{ {R}_{eq} = R_1 + R_2}$

$\longrightarrow \sf{ r_1 = R_1 + R_2 \dots (i) }$

$\sf{\\}$

$\underline{\underline{\textbf{ \red{In parallel}:- }}}$

$\sf{ \dfrac{1}{{R}_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }$

$\longrightarrow \sf{ \dfrac{1}{{R}_{eq}} = \dfrac{R_1 + R_2}{R_1 R_2}}$

$\longrightarrow \sf{ r_2 = \dfrac{R_1 R_2}{R_1 + R_2}}$

$\longrightarrow \sf{r_2 = \dfrac{R_1 R_2}{r_1} \quad [From\:(i)]}$

$\longrightarrow \sf{ R_1 R_2 = r_1 r_2 \dots (ii)}$

$\sf{\\}$

Now,

$\sf{ {(R_1 + R_2)}^{2} = {R}_{1}^{2} + {R}_{2}^{2} + 2R_1 R_2}$

$\longrightarrow \sf{ {r}_{1}^{2} = {R}_{1}^{2} + {R}_{2}^{2} + 2r_1r_2 \quad[From\:(i)\:and\:(ii)] }$

$\longrightarrow \sf{ {R}_{1}^{2} + {R}_{2}^{2} = {r}_{1}^{2} - 2r_1 r_2\dots(iii)}$

$\sf{\\}$

$\sf{ {(R_1 - R_2)}^{2} = {R}_{1}^{2} + {R}_{2}^{2} - 2R_1 R_2}$

$\longrightarrow \sf{ {(R_1 - R_2)}^{2} = {r}_{1}^{2} - 2r_1r_2 - 2r_1 r_2\quad[From(i) and (iii)}$

$\longrightarrow \sf{ {(R_1 - R_2)}^{2} = {r}_{1}^{2} - 4r_1 r_2}$

Considering only positive values,

$\longrightarrow \sf{ R_1 - R_2 = \sqrt{ {r}_{1}^{2} - 4r_1 r_2 } \dots (iv)}$

$\sf{\\}$

$\underline{\underline{\textbf{ \red{ (i) + (iv)}:- }}}$

$\sf{ R_1 + R_2 + R_1 - R_2 = r_1 + \sqrt{ {r}_{1}^{2} - 4r_1 r_2 }}$

$\sf{R_1 =\dfrac{1}{2}(r_1 + \sqrt{ {r}_{1}^{2} - 4r_1 r_2 })}$

$\sf{\\}$

$\underline{\underline{\textbf{ \red{ (i) - (iv)}:- }}}$

$\longrightarrow \sf{R_1 + R_2 - R_1 + R_2 = r_1 - \sqrt{ {r}_{1}^{2} - 4r_1 r_2 }}$

$\longrightarrow \sf{R_2 = \dfrac{1}{2}(r_1 - \sqrt{ {r}_{1}^{2} - 4r_1 r_2 })}$

$\sf{\\}$

So,

$\sf{\dfrac{R_1}{R_2} = \dfrac{\dfrac{1}{2}(r_1 + \sqrt{ {r}_{1}^{2} - 4r_1 r_2 })}{\dfrac{1}{2}(r_1 - \sqrt{ {r}_{1}^{2} - 4r_1 r_2 })}}$

$\longrightarrow \sf{\dfrac{R_1}{R_2} = \dfrac{r_1 + \sqrt{ {r}_{1}^{2} - 4r_1 r_2 }}{r_1 - \sqrt{ {r}_{1}^{2} - 4r_1 r_2 }}}$

$\sf{\\}$

Hence, the answer is,

$\longrightarrow \underline{\underline{\sf{ \green{D)\:\dfrac{R_1}{R_2} = \dfrac{r_1 + \sqrt{ {r}_{1}^{2} - 4r_1 r_2 }}{r_1 - \sqrt{ {r}_{1}^{2} - 4r_1 r_2 }} }}}}$

Answer 2

According to the question,

$\sf{\longrightarrow r_2=\dfrac{R_1R_2}{R_1+R_2}}$

Multiply both sides by 4.

$\sf{\longrightarrow 4r_2=\dfrac{4R_1R_2}{R_1+R_2}}$

We know that,

• $\sf{(a+b)^2-(a-b)^2=4ab}$

Therefore, on taking $\sf{a=R_1}$ and $\sf{b=R_2,}$

$\sf{\longrightarrow 4r_2=\dfrac{(R_1+R_2)^2-(R_1-R_2)^2}{R_1+R_2}}$

Dividing both sides by $\sf{R_1+R_2,}$

$\sf{\longrightarrow\dfrac{4r_2}{R_1+R_2}=\dfrac{(R_1+R_2)^2-(R_1-R_2)^2}{(R_1+R_2)^2}}$

$\sf{\longrightarrow\dfrac{4r_2}{R_1+R_2}=1-\left(\dfrac{R_1-R_2}{R_1+R_2}\right)^2}$

But according to the question,

$\sf{\longrightarrow r_1=R_1+R_2}$

Thus,

$\sf{\longrightarrow\dfrac{4r_2}{r_1}=1-\left(\dfrac{R_1-R_2}{R_1+R_2}\right)^2}$

$\sf{\longrightarrow\left(\dfrac{R_1-R_2}{R_1+R_2}\right)^2=1-\dfrac{4r_2}{r_1}}$

$\sf{\longrightarrow\dfrac{R_1-R_2}{R_1+R_2}=\dfrac{\sqrt{r_1-4r_2}}{\sqrt{r_1}}}$

Rationalising the RHS, by multiplying both numerator and denominator with $\sf{\sqrt{r_1},}$

$\sf{\longrightarrow\dfrac{R_1-R_2}{R_1+R_2}=\dfrac{\sqrt{r_1\,\!^2-4r_1r_2}}{r_1}}$

Finally, by rule of componendo and dividendo,

$\sf{\longrightarrow\underline{\underline{\dfrac{R_1}{R_2}=\dfrac{r_1+\sqrt{r_1\,\!^2-4r_1r_2}}{r_1-\sqrt{r_1\,\!^2-4r_1r_2}}}}}$

Hence (D) is the answer.