In the circuit shown,the resistance r is a va | vedclass

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(C) The circuit consists of a battery of voltage $V$ connected in series with a resistor $R$ and a parallel combination of $R$ and $r$.The equivalent

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$\frac{1}{2}$

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(C) The circuit consists of a battery of voltage $V$ connected in series with a resistor $R$ and a parallel combination of $R$ and $r$.The equivalent resistance of the parallel part is $R_p = \frac{Rr}{R+r}$.The total resistance of the circuit is $R_{eq} = R + \frac{Rr}{R+r} = \frac{R^2 + 2Rr}{R+r}$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{V(R+r)}{R(R+2r)}$.The current through the variable resistor $r$ is given by the current divider rule: $I_r = I 	imes \frac{R}{R+r} = \frac{V(R+r)}{R(R+2r)} 	imes \frac{R}{R+r} = \frac{V}{R+2r}$.The heat generation in $r$ is $H = I_r^2 r = \left(\frac{V}{R+2r}ight)^2 r = \frac{V^2 r}{(R+2r)^2}$.To find the maximum heat,we differentiate $H$ with respect to $r$ and set it to zero: $\frac{dH}{dr} = V^2 \left[ \frac{(R+2r)^2 - r \cdot 2(R+2r) \cdot 2}{(R+2r)^4} ight] = 0$.This implies $(R+2r)^2 - 4r(R+2r) = 0$.Since $R+2r 
eq 0$,we have $R+2r - 4r = 0$,which gives $R - 2r = 0$,or $r = \frac{R}{2}$.Given $r = fR$,we find $f = \frac{1}{2}$.

In the circuit shown,the resistance $r$ is a variable resistance. If for $r = fR$,the heat generation in $r$ is maximum,then the value of $f$ is:

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