Question

When two unknown resistors are connected in series with a battery, 225 W power is delivered to the combination with a total current of 5 A. For the same resistors, total power 50 W is delivered when they are connected in parallel. Find the ratio of the two resistances.

Answer 1

Here power delivered in series and parallel combinations is
In series, $P=5^2(R_1+R_2)$
$225=5^2(R_1+R_2)$ .........(1)
In parallel, $P=I^2\left(\frac{R_1{R}_2}{R_1+R_2}\right)$
$50=5^2\left(\frac{R_1{R}_2}{R_1+R_2}\right)$ ..........(2)
comparing equations (1) and (2), we get
$\frac{(R_1+R_2{)}^2}{R_1{R}_2}=\frac{225}{50}=\frac{9}{2}$
$\frac{R_1^2(1+\frac{R_2}{R_1}{)}^2}{R_1{R}_2}=\frac{9}{2}$
Put, $\frac{R_1}{R_2}=x$
$x(1+\frac{1}{x}{)}^2=\frac{9}{2}$
$x(\frac{x+1}{x}{)}^2=\frac{9}{2}$
$x^2+2x+1=\frac{9x}{2}$
$2x^2-5x+2=0$
Taking roots of equation we get
$x=2$
$\therefore \frac{R_1}{R_2}=2$