The current measuring capacity of a galvanometer of resistance `50 Omega` is to be increased from x mA to `101 xxmA`.The shunt resistance is

The combined resistance of a galvanometer of resistance 495 Omega and its shunt is 9.9 Omega . Find the shunt resistance.

The maximum current that can be measured by a galvanometer of resistance 40 Omega is 10 mA. It is converted into a voltmeter that can read upto 50 V. The resistance to be connected in the series with the galvanometer is ............ (in ohm).

The maximum current that can be measured by a galvanometer of resistance 40 Omega is 10 mA . It is converted into a voltmeter that can read upto 50 V . The resistance to be connected in series with the galvanometer is ... (in ohm )

If a shunt is to be applied to a galvanometer of resistance 50Omega so that only 5% of total current passes through the galvanometer. The resistance of shunt should be

In an ammeter, 4% of the total current is passing through the galvanomter. If the shunt resistance is 5 Omega , then the resistance is Omega , then the resistance of the galvanometer will be

In the given circuit, a galvanometer with a resistance of 70 Omega is converted to an ammeter 0.05 Omega by a shunt resistance of 0.05 2, total current measured by this device

In the given circuit, a galvanometer with a resistance of 70 Omega is converted to an ammeter 0.05 Omega by a shunt resistance of 0.052, total current measured by this device

Current is sent through a galvanometer of resistance 50 Omega from a battery of emf 30V and of negligible internal resistance.There is a shunt of resistance 2 Omega connected in parallel with the galvanometer.Find out the resistance that should be connected in series with the battery to get a current of 10mA through the galvanometer.