In a potentiometer experiment, the galvanometer shows no deflection when a cell is connected across $60\, cm$ of the potentiometer wire. If the cell is shunted by a resistance of $6\,\Omega $, the balance is obtained across $50\, cm$ of the wire. The internal resistance of the cell is .............. $\Omega $
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Temperature coefficient at $0\,^oC$ is $0.00125\,^oC^{-1}$. At a temperature of $25\,^oC$ its resistance is $1\,\Omega $. Find the temperature at which resistance is $1.2\,\Omega $
It is preferable to measure the $e.m.f.$ of a cell by potentiometer than by a voltmeter because of the following possible reasons.
$(i)$ In case of potentiometer, no current flows through the cell.
$(ii)$ The length of the potentiometer allows greater precision.
$(iii)$ Measurement by the potentiometer is quicker.
$(iv)$ The sensitivity of the galvanometer, when using a potentiometer is not relevant.
Which of these reasons are correct?
We have a galvanometer of resistance $25\,\Omega $. It is shunted by a $2.5\,\Omega $ wire. The part of total current that flows through the galvanometer is given as
The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as $R_1$ has the colour code (Orange, Red, Brown). The resistors $R_2$ and $R_4$ are $80\, \Omega $ and $40\,\Omega $, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as $R_3$ would be
A wire of resistance $R_{1}$ is drawn out so that its length is increased by twice of its original length.The ratio of new resistance to original resistance is.
When an ammeter of negligible internal resistance is inserted in series with circuit it reads $1A$. When the voltmeter of very large resistance is connected across $X$ it reads $1V$. When the point $A$ and $B$ are shorted by a conducting wire, the voltmeter measures $10\, V$ across the battery. The internal resistance of the battery is equal to .............. $\Omega$
In the adjoining circuit, the battery ${E_1}$ has an $e.m.f.$ of $12\,volt$ and zero internal resistance while the battery $E$ has an $e.m.f.$ of $2\,volt$. If the galvanometer $G$ reads zero, then the value of the resistance $X$ in $ohm$ is