Kindly give me correct answer only. Its urgent. Plz help me! The question is as follows; (Derive the principle of wheatstone bridge?)
Answers
Answer:
Wheatstone bridge principle states that if four resistances P, Q, R and S are arranged to form a bridge with a cell E and one way key, put between the points A and C and a galvanometer is connected in between the points B and D such that there is no current through G.
Answer:
The Wheatstone bridge works on the principle of null deflection, i.e. the ratio of their resistances are equal and no current flows through the circuit. Under normal conditions, the bridge is in the unbalanced condition where current flows through the galvanometer. The bridge is said to be in a balanced condition when no current flows through the galvanometer. This condition can be achieved by adjusting the known resistance and variable resistance.
The current enters the galvanometer and divides into two equal magnitude currents as I1 and I2. The following condition exists when the current through a galvanometer is zero,
I1P=I2R (1)
The currents in the bridge, in a balanced condition, is expressed as follows:
I1=I3=EP+Q I2=I4=ER+S
Here, E is the emf of the battery.
By substituting the value of I1 and I2 in equation (1), we get
PEP+Q=RER+S PP+Q=RR+S P(R+S)=R(P+Q) PR+PS=RP+RQ PS=RQ (2)
R=PQ×S (3)
Equation (2) shows the balanced condition of the bridge while (3) determines the value of the unknown resistance.
In the figure, R is the unknown resistance, and the S is the standard arm of the bridge and the P and Q are the ratio arm of the bridge.
Explanation:
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