Questions in Current Electricity

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	The resistance of 10 metre long potentiometer wire is 1ohm/meter. A cell of e.m.f.2.2 volts and a high resistance box are connected in series to this wire. The value of resistance taken from resistance box for getting potential gradient of 2.2 millivolt/metre will be
	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
	In the adjoining circuit, the e.m.f. of the cell is 2 volt and the internal resistance is negligible. The resistance of the voltmeter is 80 ohm. The reading of the voltmeter will be
	If the resistivity of a potentiometer wire be $\rho$ and area of cross-section be A, then what will be potential gradient along the wire
	A voltmeter has resistance of 2000 ohms and it can measure upto 2V. If we want to increase its range to 10 V, then the required resistance in series will be
	For a cell of e.m.f. 2V, a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a $2\,\Omega $ resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is
	The arrangement as shown in figure is called as
	A potentiometer wire of length $1\,m$ and resistance 10 $\Omega$ is connected in series with a cell of emf 2V with internal resistance 1 $\Omega$ and a resistance box including a resistance R. If potential difference between the ends of the wire is 1 mV, the value of R is
	In a balanced Wheatstone’s network, the resistances in the arms Q and S are interchanged. As a result of this
	The ammeter A reads 2 A and the voltmeter V reads 20 V. the value of resistance R is (Assuming finite resistance's of ammeter and voltmeter)

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