A charged capacitor is allowed to dischare through a resistor by closing the key at the instant `t=0`. At the instant `t=(ln 4)mus`, the reading of the ammeter falls half the initial vaslue. The resistance of the ammeter is equal to

A charged capacitor is allowed to discharged through a resistor 2 Omega by closing the switch K at the t=0 . At time t = 2 mus , the reading of the falls half of its initial value. The resistance of ammeter is equal to

A charged capacitor is allowed to discharged three a resistor 2 Omega by closing the switch S at the t=0 . At time t = 2 mus , the reading of the falls half of its initial value. The resistance of ammeter is equal to

In the circuit shown the key (K) is closed at t=0 , the current through the key at the instant t=10^-3 ln, 2 is

In the circuit shown the key (K) is closed at t=0 , the current through the key at the instant t=10^-3 ln, 2 is

An ammeter and a voltmeter are initially connected in series to a battery of zero internal resistance. When switch S_(1) is closed the reading of the voltmeter becomes half of the initial, whereas the reading of the ammeter becomes double. If now switch S_(2) is also closed, then reading of ammeter becomes :

An ammeter and a voltmeter are initially connected in series to a battery of zero internal resistance. When switch S_(1) is closed the reading of the voltmeter becomes half of the initial, whereas the reading of the ammeter becomes double. If now switch S_(2) is also closed, then reading of ammeter becomes :

An initially uncharged capacitor C is being charged by a battery of emf E through a resistance R upto the instant when the capacitor is charged to the potential E/2, the ratio of the work done by the battery to the heat dissipated by the resistor is given by :-

The figure-3.375 shows two circuits with a charged capacitor that is to be discharged through a resistor as shown in the figure. The ratio of initial charges on capacitors is given as q_(2)//q_(1) = 2 . If both switches are closed at time t= 0 , the charges become equal at 10^(-4) ln 2 s . Find the resistance R.

In the circuit shows in Fig. kay K is closed at t = 0 , the current through the key at the instant t = 10^(-3) In 2 s is