A current source drives a current in a coil o | vedclass

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Question

Let internal resistance of source $=\mathrm{R}$

Current in coil of resistance

$\mathrm{R}_{1}=\mathrm{I}_{1}=\frac{\mathrm{V}}{\mathrm{R}+\mathr

Vedclass · Accepted Answer

$\sqrt {R_1R_2}$

Vedclass · Answer

Let internal resistance of source $=\mathrm{R}$

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 $\quad\left(\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{1}}\right)^{2} \mathrm{R}_{1}=\left(\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{2}}\right)^{2} \mathrm{R}_{2}$

$\Rightarrow \quad \mathrm{R}_{1}\left(\mathrm{R}+\mathrm{R}_{2}\right)^{2}=\mathrm{R}_{2}\left(\mathrm{R}+\mathrm{R}_{1}\right)^{2}$

$\Rightarrow \mathrm{R}^{2} \mathrm{R}_{1}+\mathrm{R}_{1} \mathrm{R}_{2}^{2}+2 \mathrm{RR}_{1} \mathrm{R}_{2}$

$\Rightarrow \quad \mathrm{R}^{2} \mathrm{R}_{2}+\mathrm{R}_{1}^{2} \mathrm{R}_{2}+2 \mathrm{RR}_{1} \mathrm{R}_{2} ?$

$\Rightarrow \quad \mathrm{R}^{2}\left(\mathrm{R}_{1}-\mathrm{R}_{2}\right)=\mathrm{R}_{1} \mathrm{R}_{2}\left(\mathrm{R}_{1}-\mathrm{R}_{2}\right)$

$\Rightarrow \quad \mathrm{R}=\sqrt{\mathrm{R}_{1} \mathrm{R}_{2}}$

A current source drives a current in a coil of resistance $R_1$ for a time $t$. The same source drives current in another coil of resistance $R_2$ for same time. If heat generated is same, find internal resistance of source.

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