The current through the solenoid is changing in such way that flux through it is given by `phi=varepsilon` t .The solenoid is surrounded by a loop having resistance `R_(1)` and `R_(2)` as shown.Then the reading of the two voltmeters `V_(1)` and `V_(2)` differ by :

In the circuit shown, the readings of voltmeters V_(1), V_(2) and V_(3) are given by

If the resistance R_(1) is increased how will the reading of the ammeter and voltmeter change

A solenoid is at potential difference 60 V and current flows through it is 15 ampere , then the resistance of coil will be

A voltmeter of resistance 600 Omega when connected in turn across resistances R_(1) and R_(2) gives readings of V_(1) and V_(2) , respectively. If the battery is ideal, then

An ammeter A, a voltmeter V and a resistance R are connected as shown in the figure. If the voltmeter reading is 1.6V and the ammeter reading is 0.4 A, then R is

The ammeter A reads 2A and the voltmeter V reads 20V, the value of resistance R is (Assuming finite resistance's of ammeter and voltmeter)

In the circuit shown in figure-3.391, V_(1) and V_(2) are two voltmeters having resistances 6000 Omega and 4000 Omega respectively EMF of battery is 250V, having negligible internal resistance. Two resistances R_(1) and R_(2) are 4000 Omega and 6000 Omega respectively. Find the reading of the voltamter V_(1) and V_(2) when (a) Switch S is open ltbgt (b) Switch S is closed

Two ideal batteries of emf V_(1) and V_(2) and three resistance R_(1)R_(2) and R_(3) are connected The current in resistance R_(2) would be zero if