Current Electri...

The net resistance of a voltmeter should be large to ensure that

A potentiometer circuit is set up as shown. The potential gradient, across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths l₁ cm and 1₂ cm respectively. The magnitudes, of theresistors R and X, in ohms, are then, equal, respectively,to

A potentiometer consists of a wire of length 4 m and resistance 10 Ω. It is connected to a cell of e.m.f. 2 V. The potential difference per unit length of the wire will be

In a meter bridge, the balancing length from the left end (standard resistance of one ohm is in the right gap) is found to be 20 cm. The value of the unknownresistance is

In the circuit shown P ≠ R, the reading of the galvanometer is same with switch S open or closed. Then

In the following Wheatstone bridge P/Q = R/S. If keyK is closed, then the galvanometer will show deflection

A galvanometer having a resistance of 8 ohm is shunted by a wire of resistance 2 ohm. If the total current is 1 amp, the part of it passing through the shunt will be

A galvanometer whose resistance is 120 Ω gives full scale deflection with a current of 0.05 A so that it can read a maximum current of 10 A. A shunt resistance is added in parallel with it. The resistance of the ammeter so formed is

A voltmeter of resistance 1000 Ω gives full scale deflection when a current of 100 mA flows through it. The shunt resistance required across it to enable it to be used as an ammeter reading 1 A at full scale deflection is

We have a galvanometer of resistance 25 Ω. It is shunted by a 2.5 Ω wire. The part of total current that flows through the galvanometer is given as