The sensitivity of potentiometer can be increased by decreasing the potential gradient i.e., by increasing the length of potentiometer wire. (Sensitivity $\propto \frac{1}{P . G .}$ and Length)
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A cable of resistance $10\,\Omega $ carries electric power from a generator producing $250\, kW$ at $10000\, volt$, the power lost in the cable during transmission is ............. $kW$
In the given circuit of potentiometer, the potential difference $E$ across $AB$ ( $10\, m$ length) is larger than $E _{1}$ and $E _{2}$ as well. For key $K _{1}$ (closed), the jockey is adjusted to touch the wire at point $J_{1}$ so that there is no deflection in the galvanometer. Now the first battery $\left( E _{1}\right)$ is replaced by second battery $\left( E _{2}\right)$ for working by making $K _{1}$ open and $K _{2}$ closed. The galvanometer gives then null deflection at $J _{2}$. The value of $\frac{ E _{1}}{ E _{2}}$ is $\frac{ a }{ b },$ where $a =$ ...............
Every atom makes one free electron in copper. If $1.1$ $ampere$ current is flowing in the wire of copper having $1\, mm$ diameter, then the drift velocity (approx.) will be (Density of copper $ = 9 \times {10^3}\,kg\,{m^{ - 3}}$ and atomic weight = $63$)
An ammeter $A$ of finite resistance, and a resistor $R$ are joined in series to an ideal cell $C$. $A$ potentiometer $P$ is joined in parallel to $R$. The ammeter reading is $I_0$ and the potentiometer reading is $V_0$. $P$ is now replaced by a voltmeter of finite resistance. The ammeter reading now is $I$ and the voltmeter reading is $V$.