Post Office Box Experiment Viva Questions

Post Office Box Experiment Viva Questions with Answers

(i) What is the principle of Post Office Box?

Ans: Wheatstone principle P/Q = R/S

(ii) What is the wire used in the resistance box?

Ans: Manganin or constantan

(iii) Does current flow through the galvanometer at the null point?

Ans: The current through the galvanometer is zero

(iv) Why do you press battery key first and then galvanometer key?

Ans: If you press the galvanometer first and then battery key there is induced current in the circuit and is difficult to adjust the balancing

(v) What is temperature coefficient of resistance?

Ans: Temperature coefficient, α = Rt – R0 / R0 x t

(vi) What is meant by negative temperature coefficient?

Ans: The resistance decreases with increase of temperature. Semiconductors have negative temperature coefficient

(vii) Aim of the Post Office Box Experiment?

Ans:
(a)To determine the resistivity of the material of a wire
(b)To verify the laws of resistances

(viii) Apparatus of the Post Office Box Experiment?

Ans: Post office box, resistance wire, battery, galvanometer, battery etc.

(ix)Theory of the Post Office Box Experiment?

Ans: For a balanced Wheatstone's bridge,

P/ Q = R/ X; where P and Q are the resistances in the ratio arms, R is the resistance in the resistance arm and X is the resistor of unknown resistance. If r is the radius and l is the length of X, its resistivity,

ρ = X(πr2/l)

(x) Procedure for determining the resistivity of the material of a wire

Ans: The electrical connection are made as in the figure. Resistances of 10 ohm each are intro-duced in the ratio arms P and Q. Keeping the resistance in the resistance arm R zero, the tap keys K1 and K2 are closed. The direction of deflection is noted. The infinity plug in the resistance arm is now removed. The keys are again pressed. If the deflections are in the opposite directions, the connections are correct. The infinity plug is now inserted. Resistances are introduced in the arm R till the galvanometer shows opposite deflections for two resistances differing by one ohm. Let these resistances be 2 and 3 ohm. According to Wheatstone' principle,

P/Q = R/X = 1

Therefore, X = R. Hence X also lies between 2 and 3 ohm.

Then 100 ohms resistance is taken in the arm P so that the ratio P/ Q = 10 = R/ X;

Therefore, X = R/ 10. As before two resistances differing by one ohm are taken in R so as to get opposite deflections. Let them be 24 and 25 ohm. So X lies between 2.4 and 2.5 ohm. Then 1000 ohms is introduced in R so that P/Q = 100 = R/X;

Therefore, X = R/100.

If 245 and 246 ohms are the resistances to be taken in R so as to get opposite deflections, then X lies between 2.45 and 2.46 ohm. Usually the mean resistances 2.455 ohm is taken as the value of X. The length L and the radius r of the wire are measured. Resistivity is calculated using the formula,
ρ = Xπr2/L

(xi) Procedure for verifying the laws of resistances

Ans:
(a) Resistances in series : The two wires are joined in series and the resistance X3 of the combination is determined. It is found that,
X3 = X1 + X2; (approximately)

(b) Resistance in parallel : The two wires are joined in parallel and the resistance X4 of the combination is determined. It is found that,
X4 =   X1X2/ X1+X2 ; (approximately). Hence Verified.

0 Comments