In the circuit shown below (on the left) the resistance and the emf source are both variable. The graph of seven readings of the voltmeter and the ammeter ( $V$ and $I$, respectively) for different settings of resistance and the emf, taken at equal intervals of time $\Delta t$, are shown below (on the right) by the dots connected by the curve $E F G H$. Consider the internal resistance of the battery to be negligible and the voltmeter an ammeter to be ideal devices. (Take, $R_0 \equiv \frac{V_0}{I_0}$ ).
Then, the plot of the resistance as a function of time corresponding to the curve $E F G H$ is given by
KVPY 2017, Advanced
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(d)
In given $V-I$ graph,
From $E$ to $F$,
$V_0=I_0 R_0$
$\Rightarrow \quad \text { s lope }=R_0$
$\therefore \text { At } F, \text { resistance }=R_0=\frac{V_0}{I_0}$
From $F$ to $G$,
$V=V_0=$ constant
But current increases, so resistance must decreases.
At $G$, resistance $=\frac{R_0}{2}=\frac{V_0}{2 I_0}$.
From $G$ to $H$, current is constant but voltage increases, so resistance decreases.
At $H$, resistance $=\frac{2 V_0}{2 I_0}=\frac{V_0}{I_0}=R_0$
So, correct option is $(d)$.
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