Voltage required $=i R=1 \times 10=10 \mu V$
$\therefore 40 \mu V \cong 1^{\circ} C$
$10 \mu V \cong \frac{1}{(4)^{\circ}} C$
$=2.50^{\circ} C$



$(A)$ the current $I$ through the battery is $7.5 \mathrm{~mA}$
$(B)$ the potential difference across $R_{\mathrm{L}}$, is $18 \mathrm{~V}$
$(C)$ ratio of powers dissipated in $R_1$ and $R_2$ is $3$
$(D)$ if $R_1$ and $R_2$ are interchanged, magnitude of the power dissipated in $R_{\mathrm{L}}$ will decrease by a factor of $9$