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

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 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 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

In the circuit shown, R=2 Omega and V = 20 volt. With switch S open the reading of ammeter is half its reading when ‘S’ is closed. Calculate the resistance of the ammeter.

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t_1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t_2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t_1//t_2 will be

LetC be the capacitance of a capacitor discharging through a resistor R. Suppose t_1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t_2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t_1//t_2 will be

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 :