91Ó°ÊÓ

The resistances when connected in series, the equivalent resistance is,

$R_{eq}=R_1+R_2$

If the voltage applied is É›, then the current in the circuit is,

$I=\frac{\mathrm{Î\mu }}{R_1+R_2}$

Thus, the power dissipated is,

$\begin{array}{r}{P}_s=I^2\left(R_1+R_2\right)\\ ={\left[\frac{\mathrm{Î\mu }}{R_1+R_2}\right]}^2\left(R_1+R_2\right)\\ =\frac{{\mathrm{Î\mu }}^2}{\left(R_1+R_2\right)}\end{array}$

Substitute all the values in the above equation,

$$\begin{gathered} 48.0 W=\frac{(48.0 V{)}^2}{(R_1+R_2)} \\ {R}_1+R_2=48.0 \mathrm{Î\copyright } \end{gathered}$$ ...(i)

The resistances when connected in parallel, the equivalent resistance is,

$R_{eq}=1/R_1+1/R_2$

If the voltage applied is É›, then the current in the circuit is,

$I=\frac{\mathrm{Î\mu }}{1/R_1+1/R_2}$

Thus, the power dissipated is,

$\begin{array}{r}{P}_p=l_1^2{R}_1+l_2^2{R}_2\\ =\frac{{\mathrm{Î\mu }}^2}{R_1^2}{R}_1+\frac{{\mathrm{Î\mu }}^2}{R_2^2}{R}_2\\ ={\mathrm{Î\mu }}^2\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\\ ={\mathrm{Î\mu }}^2\left(\frac{R_1+R_2}{R_2{R}_1}\right)\\ =256\mathrm{W}\end{array}$

Substitute all the values in the above equation,

$\begin{array}{r}(48.0\mathrm{V}{)}^2\frac{R_1{R}_2}{\left(R_1+R_2\right)}=256\mathrm{W}\\ {\text{Use,}}_1{R}_1+R_2=48.0\mathrm{Î\copyright }\\ R_1{R}_2=432{\mathrm{Î\copyright }}^2\end{array}$ ...(ii)

Solve equation (i) and (ii) for both the resistances to get two sets of answers,

$\begin{array}{r}{R}_1=36.0\mathrm{Î\copyright }\text{and}{R}_2=12.0\mathrm{Î\copyright }\\ R_1=12.0\mathrm{Î\copyright }\text{and}{R}_2=36.0\mathrm{Î\copyright }\end{array}$

Since, it is given in question that R₁ > R₂. Thus the first pair is correct. Thus,$R_1=36.0\mathrm{Î\copyright } and R_2=12.0\mathrm{Î\copyright }$

In the series combination,

The same current flows through the resistances. As the power is given as $\mathrm{P}={\mathrm{I}}^2\mathrm{R}$, so more is the resistance more the power consumed. As established R₁ > R₂ , therefore more power will be lost at R₁.

In the parallel combination,

The same current does not flow through the resistances but the voltage is same. As the power is given as $\mathrm{P}={\mathrm{V}}^2/\mathrm{R}$, so more is the resistance less is the power consumed. As established R₁ > R₂ , therefore less power will be lost at R₁.