Current in coil of resistance
$\mathrm{R}_{1}=\mathrm{I}_{1}=\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{1}}$
Current in coil of resistance
$\mathrm{R}_{2}=\mathrm{I}_{2}=\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{2}}$
Further, as heat generated is same, so
$\mathrm{I}_{1}^{2} \mathrm{R}_{1} \mathrm{t}=\mathrm{I}_{2}^{2} \mathrm{R}_{2} \mathrm{t}$
or $\quad\left(\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{1}}\right)^{2} \mathrm{R}_{1}=\left(\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{2}}\right)^{2} \mathrm{R}_{2}$
$\Rightarrow \quad \mathrm{R}_{1}\left(\mathrm{R}+\mathrm{R}_{2}\right)^{2}=\mathrm{R}_{2}\left(\mathrm{R}+\mathrm{R}_{1}\right)^{2}$
$\Rightarrow \mathrm{R}^{2} \mathrm{R}_{1}+\mathrm{R}_{1} \mathrm{R}_{2}^{2}+2 \mathrm{RR}_{1} \mathrm{R}_{2}$
$\Rightarrow \quad \mathrm{R}^{2} \mathrm{R}_{2}+\mathrm{R}_{1}^{2} \mathrm{R}_{2}+2 \mathrm{RR}_{1} \mathrm{R}_{2} ?$
$\Rightarrow \quad \mathrm{R}^{2}\left(\mathrm{R}_{1}-\mathrm{R}_{2}\right)=\mathrm{R}_{1} \mathrm{R}_{2}\left(\mathrm{R}_{1}-\mathrm{R}_{2}\right)$
$\Rightarrow \quad \mathrm{R}=\sqrt{\mathrm{R}_{1} \mathrm{R}_{2}}$
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${V}_{{m}}({t})=10 \sin \left(2 \pi \times 10^{5} {t}\right)$ $volts$ and
Carrier signal
${V}_{{c}}({t})=20 \sin \left(2 \pi \times 10^{7} {t}\right)$ $volts$
The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is $\alpha\, {kHz}$. The value of $\alpha$ is :
