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State the principle of the Potentiometer.
A potentiometer is also called a voltmeter of infinite resistance, why?
State the two Kirchoff's laws for electrical circuits.
A cell has an emf of 1.5 V. When short circuited, it gives a current of 3 A. The internal resistance of the cell is
In the given circuit current $\rm I_1$ and $\rm I_2$ are:
Four resistors P, Q, R and S, having resistance 2, 2, 2 and 3 $\rm \Omega $ respectively, are arranged to form a wheatstone bridge. The value of the resistance with which S must be shunted in order to balance the bridge is
A 2.0 V potentiometer is used to determine the internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 75 cm. When a resistor of 10 $\Omega$ is connected across the cell, the balance point shifts to 60 cm. The internal resistance of the cell is :
A potentiometer wire is 10 m long and has a resistance of $\rm 20 \;\Omega $. It is connected in series with a battery of emf 3 V and a resistance of $ \rm 10 \Omega $. The potential gradient along the wire in V/m is :
Kirchhoff's second law is based on the law of conservation of:
Kirchoff's second law is based on the law of conservation of
A potentiometer wire of length 10 m and resistance 30 $\rm \Omega$ is connected in series with a battery of emf 2.5 V with internal resistance 5 $\rm \Omega$ and an external resistance R. If the fall of potential along the potentiometer wire is 50 mV/m, the value of R is (in $\rm \Omega$)
Electromotive force is most closely related to
Potentiometer measures potential more accurately because
A car battery has e.m.f. 12 V and internal resistance $\rm 5\; $x$\; 10^{-2}\; \Omega $. If it draws 60 A current, the terminal voltage of the battery will be
An ammeter and a voltmeter of resistance R are connected in series to an electric cell of negligible internal resistance. Their reading are A and V respectively. If another resistance R is connected in parallel with the voltmeter.
In a typical Wheatstone network, the resistance in cyclic order are P = 10 $\Omega$, Q = 5 $\Omega$, S = 4 $\Omega$, and R = 4 $\Omega$. For the bridge to balance
A battery is connected in series with a resistance R and a voltmeter. An ammeter is a connected in parallel with the battery.
A milliammeter of range 10 mA has a coil of resistance 1 $\Omega$. To use it as a voltmeter of range 10 V, the resistance that must be connected in series with it is
A battery of emf 10 V and internal resistance 3 $\Omega$ is connected to a resistor. The current in the circuit is 0.5 A. The terminal voltage of the battery when the circuit is closed is
A voltmeter having a resistance of $\rm 50\;$x$\;10^3 \;\Omega$ is used to measure the voltage in a circuit. To increase the range of measurement 3 times the additional series resistance required is:
To send 10% of the main current through a moving coil galvanometer of resistance 99 $\rm \Omega$, the shunt required is
For resistances of 100 $\Omega$ each are connected in the form of a square. Then the equivalent resistance between the diagonally opposite points is:
When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 A. When the resistance is increased to 5 ohm, the current is 0.25 A. The e.m.f of the cell is: