Potentiometer Questions

A cell can be balanced against $110\text{cm}$ and $100\text{cm}$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10\Omega$. Its internal resistance is:

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is $k\text{ volt/cm}$ and the ammeter reads $1.0\text{ A}$ when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths $l_1\text{ cm}$ and $l_2\text{ cm}$ respectively. The magnitudes, of the resistors $R$ and $X$, in ohms, are:

A potentiometer circuit has been set up for finding the internal resistance of a given cell. When the resistance, $R$, connected across the given cell, has values of (i) infinity and (ii) $9.5\Omega$, the 'balancing lengths' are found to be $3\text{m}$ and $2.85\text{m}$, respectively. The value of internal resistance of the cell is:

A potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given by:

A potentiometer wire is $100\text{cm}$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50\text{cm}$ and $10\text{cm}$ from the positive end of the wire in the two cases. The ratio of emf’s is: