In a wheatstone bridge P=10 ohm , Q=20 ohm, R=30 ohm. the value of S in order that no current may flow through the galvanometer is

The resistance of a galvanometer is 90 ohm s. If only 10 percent of the main current may flow through the galvanometer, in which way and of what value, a resistor is to be used

In the circuit shown , the value of R in ohm that will result in no current through the 30 V battery is

All the resistance in the given Wheatstone bridge have different values and the current through the galvanometer is zero. The current through the galvanometer will still be zero if,

A galvanometer of resistance 150 ohm is shunted such that only 1/11 of the main current flows through the galvanometer. The resistance of the shunt is, (in ohm)

Find current through 4 ohm resistance

In Wheatstone's bridge P = 9 ohm , Q = 11 ohm , R = 4 ohm and S = 6 ohm . How much resistance must be put in parallel to the resistance S to balance the bridge

For what value of R, (in ohm) the current in galvanometer is zero?

In the Wheatstone network given, P=10 Omega, Q=20Omega, R=15Omega, S=30Omega , the current passing through the battery (of negligible internal resistance)

The resistance of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm 30 ohm and 90 ohm rerspectively. The e.m.f. and internal resistance of the cell are 7 V and 5 ohm respectively. If the galvanometer resistance is 50 ohm , the current drawn for the cell will be

In the given circuit, the resistances are given in ohm. The current through the 10Omega resistance is 3 A while that through the resistance X is 1 A. No current passes through the galavanometer. The values of the unknown resistances X and Y are respectively (in ohm)