In a potentiometer wire experiment the $\mathrm{emf}$ of a battery in the primary circuit is $20\,V$ and its internal resistance is $5\,\Omega$ . There is a resistance box in series with the battery and the potentiometer wire, whose resistance can be varied from $120\,\Omega$ to $170\,\Omega$ . Resistance of the potentiometer wire is $75\,\Omega$ . The following potential differences can be measured using this potentiometer.
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In the circuit shown in figure reading of voltmeter is $V_1$ when only $S_1$ is closed, reading of voltmeter is $ V_2$ when only $S_2$ is closed and reading of voltmeter is $V_3$ when both $S_1$ and $S_2$ are closed. Then
If $n, e, \tau$ and $m$ are representing electron density, charge, relaxation time and mass of an electron respectively, then the resistance of a wire of length / and cross-sectional area $A$ is given by
In the adjoining circuit, the battery $E_1$ has an $e.m.f.$ of $12\, volts$ and zero internal resistance, while the battery $E_2$ has an $e.m.f.$ of $2\, volts$. If the galvanometer $G$ reads zero, then the value of the resistance $X$ in $ohms$ is
In the figure shown, what is the current (in Ampere) drawn from the battery ? You are given $R_1 = 15\,\Omega $$,R _2 = 10\,\Omega ,$$ R_3 = 20\,\Omega ,$$ R_4 = 5\,\Omega ,$$R_5 = 25\,\Omega ,$$R_6 = 30\,\Omega , $$E = 15\,V$
In a potentiometer experiment the balancing with a cell is at length $240\, cm$. On shunting the cell with a resistance of $2$ $\Omega$, the balancing length becomes $120\, cm$. The internal resistance of the cell is ................. $\Omega $
A current of $1.6\, A$ is flowing through a wire having cross-sectional area $1\, mm^2$. If density of free electrons in the material of the wire is $10^{29}\, per\, m^3$, the drift velocity of electrons will be