Obtain the balancing condition in case of a Wheatstone’s network.
Answers
Answer:
In balanced Wheatstone's bridge, zero current flow through galvanometer. Means while applying Kirchoff's law we can neglet this path. Look at the figure, No current flows through galvanometer G if circuit is balanced
Answer:
Meter Bridge consists of a 1 meter long standard resistance wire of uniform cross-section held taut between two ends of metallic strip bent at right angle at the ends. There are two gaps, one on the left side and the other on the right side, in the horizontal part of the metallic strip where two resistors, one of which is known and the other unknown are attached. The knife edge end of the jokey is moved on the wire until the galvanometer shows null reading. Let D be the position of the jokey on the wire, when null reading is obtained. Let ‘r’ be the resistance per centimetre of the standard wire. As the area of cross section of the standard wire is constant, its resistance is proportional to its length. Resistance of the wire AD = r l, Resistance of the wire DC = r(100 – l) Wheatstone Bridge and Meter Bridge circuits are completely similar. The following table gives the comparison of resistances shown in the Wheatstone Bridge and Meter Bridge. As the current in the galvanometer of the meter bridge is zero, the circuit in the meter bridge is a balanced one and hence we can apply equation 11 to the circuit. Thus we find the resistance of an unknown resistor using the principle of Wheatstone bridge in a meter bridge circuit.