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Q.
Consider the circuit shown in the figure. The value of the resistance $X$ for which the thermal power generated in it is practically independent of small variation of its resistance is
For above circuit
$\frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{X}=\frac{R+X}{R X} $
$R^{\prime}=\frac{R X}{R+X}$
The current in the circuit
$i=\frac{E}{R+R^{\prime}}=\frac{E}{\left(R+\frac{R X}{R+X}\right)}$
$V_{R X}=\frac{E \cdot \frac{R X}{R+X}}{\left(R+\frac{R X}{R+X}\right)}=\frac{E X}{R+2 X}$
Power dissipated in the circuit
$P_{X} =\frac{V_{R X}^{2}}{X}=\frac{E^{2} \cdot X^{2}}{X(R+2 X)^{2}} $
$=\frac{E^{2} X}{(R+2 X)^{2}} $
$\frac{d P_{X}}{d X} =E^{2} \frac{(R-2 X)}{(R+2 X)^{3}}$
$\Rightarrow d P_{X}=\frac{E^{2}(R-2 X)}{(R+2 X)^{3}} \cdot d X$
$\left(d P_{x}\right)$ will be zero for all $(d X)$ if
$X=\frac{R}{2} \quad\left[d P_{x}=\frac{E^{2}\left(R-2 \times \frac{R}{2}\right)}{\left(R+2 \times \frac{R}{2}\right)^{3}}\right] $
$=0$
