In given diagram value of carbon resistor is 22 × 10 5 Ω ± 5 % . Write colour of first ring A .

According to colour code of carbon resistors, red stands for 2, yellow for 4, blue for 6 and silver stands for 5%.First and second band represents the first two significant figure of resistance. Third band represents the power raised to 10 in the value of resistance. Fourth band represents the tolerance. Given carbon resistor 22 × 10 5 Ω ± 5 % thus, Red is the colour of first ring A.

Why do we get balancing point in the middle of the meter bridge generally?

The balance point is obtained in the middle of the meter bridge wire so as to increase the sensitivity of the meter bridge and there by no deflection in the galvanometer. so it minimise error in measuring resistance.

Explain the method of find unknown resistance with the help of meter bridge. Draw circuit diagram.

The key K 1 is closed to pass a current by choosing a suitable resistance R in the resistance box. The jockey is tapped along the wire to locate the null point i.e., galvanometer shows no deflection. The bridge is then said to be balanced. By principle of wheatstone's bridge, R S = I 1 100 - I 1 so, S = 100 - I 1 × R I 1 As R is known values so unknown resistance S can be calculated.

Derive formula for obtaining internal resistance of a primary cell with help of potentiometer. Draw circuit diagram.

Derivation of formula: Let the electromotive force (e.m.f.) of the experimental cell is E and its internal resistance is r. If the resistance R from the resistance box is connected with this cell then the potential difference at the ends of the cell becomes V. E = I R + r a n d V = I R then, r = E V - 1 R ......................(1) Let the potential gradient of the potentiometer be v. When R is not connected, then let the balance point is found at distance l 1 from A, then E = v I 1 .................(2) Now when R is connected with the cell, then let the balance point is obtained at a distance I 2 from A. The potential difference at the ends of the cell V = v I 2 ...............(3) Substituting equation (2) and (3) in (1) we get, r = v I 1 v I 2 - 1 R r = I 1 I 2 - 1 R

Calculate the value of unknown resistance R in given circuit. If wheatstone bridge is in balanced condition.

The bridge is balanced when, P Q = R S According to question, let A D = R ' = 1 6 + 1 R as it is in parallel combination CD=S, AB=P, BC=Q For a wheatstone network, we have four resistance P, Q, R', and S, and their values are P = 20 Ω , Q = 10 Ω , R ' = 1 6 + 1 R , S = 1 Ω 20 × 1 = 1 1 6 + 1 R × 10 R = 3 Ω

In given diagram value of carbon resistor is 22 × 10 5 Ω ± 5 % . Write colour of first ring A .

According to colour code of carbon resistors, red stands for 2, yellow for 4, blue for 6 and silver stands for 5%.First and second band represents the first two significant figure of resistance. Third band represents the power raised to 10 in the value of resistance. Fourth band represents the tolerance. Given carbon resistor 22 × 10 5 Ω ± 5 % thus, Red is the colour of first ring A.

Obtain necessary condition for balancing state of Wheat Stone Bridge by using Kirchhoff's law. Draw necessary circuit diagram.

I = 0 in galvanometer branch Applying Kirchoff's law for loop ABCD A - I 1 R 1 + 0 + I 2 R 2 = 0 I 1 R 1 = I 2 R 2 I 1 I 2 = R 4 R 1 . . . . . . . . . . ( 1 ) Applying Kirchoff's law for loop BCDB - I 1 R 3 + I 2 R 4 + 0 = 4 I 1 R 3 = I 2 R 4 I 1 I 2 = R 4 R 3 . . . . . . . . ( 2 ) from equation (1) and (2), we get, I 1 I 2 = R 4 R 3 = R 2 R 1 R 4 R 3 = R 2 R 1 R 1 R 2 = R 3 R 4 Balancing condition for wheat stone bridge