Two resistances R1 and R2 when connected in series and parallel with 120 V line, power consumed will be 25 W and 100 W, respectively. Then the ratio of power consumed by R1 to that consumed by R2 will be

Two resistances R1 and R2 when connected in series and parallel with 120 V line, power consumed will be 25 W and 100 W, respectively. Then the ratio of power consumed by R1 to that consumed by R2 will be

1. A

1 : 1

2. B

1 : 2

3. C

2 : 1

4. D

1 : 4

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Solution:

$\begin{array}{l}\mathrm{P}=\frac{{\mathrm{V}}^{2}}{\mathrm{R}}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{P}}_{\mathrm{P}}}{{\mathrm{P}}_{\mathrm{S}}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\mathrm{R}}_{\mathrm{S}}}{{\mathrm{R}}_{\mathrm{P}}}=\frac{\left({\mathrm{R}}_{1}+{\mathrm{R}}_{2}\right)}{{\mathrm{R}}_{1}{\mathrm{R}}_{2}\text{\hspace{0.17em}}/\text{\hspace{0.17em}}\left({\mathrm{R}}_{1}+{\mathrm{R}}_{2}\right)}\text{\hspace{0.17em}}=\frac{{\left({\mathrm{R}}_{1}+{\mathrm{R}}_{2}\right)}^{2}}{{\mathrm{R}}_{1}{\mathrm{R}}_{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\frac{100}{25}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{{\left({\mathrm{R}}_{1}+{\mathrm{R}}_{2}\right)}^{2}}{{\mathrm{R}}_{1}{\mathrm{R}}_{2}}\text{\hspace{0.17em}}⇒\frac{{\mathrm{R}}_{2}}{{\mathrm{R}}_{1}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{1}{1}\end{array}$

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