Question

Two resistors when connected in parallel across a cell of negligible internal resistance use fourtimes the power that they would use when connected in series across the same cell. If one of the resistors has a resistance of 10$\Omega$ the resistance of the other is

• A. 10$\Omega$
• B. 5$\Omega$
• C. 12$\Omega$
• D. 20$\Omega$

Answer 1

The correct option is A 10$\Omega$

Power in parallel $=\frac{E^2}{R_1}+\frac{E^2}{R_2}$

Power in series $=I^2[R_1+R_2]$

$=\frac{E^2}{R_1+R_2}$

$\Rightarrow \frac{4E^2}{R_1+R_2}=E^2[\frac{1}{R_1}+\frac{1}{R_2}]$

$4R_1{R}_2=(R_1+R_2{)}^2$

$\Rightarrow (R_1-R_2{)}^2=0$

$\Rightarrow R_1=R_2=10\Omega$