Draw the labelled diagram of a meter bridge. On what principle it works?

The above figure shows a meter bridge. A Meter bridge is an electrical instrument based on the Wheatstone bridge which is useful for determining the unknown resistance of any material. In this instrument, a one-meter wire is used with unknown resistance. The cross-sectional area of the wire is uniform. This wire is connected to a galvanometer. Now, as it is based on the principle of the Wheatstone bridge, we need to find the unknown resistance of the wire. The above-given figure shows the arrangement of an equivalent Wheatstone bridge. Let L be the length of the wire According to principle of Wheatstone bridge, R AD R AC = R BD R BC ⇒ R R AC = R R BC . . . . . ( 1 ) Now, we know, the relation between resistance and resistivity is given by, R = ρ L A Where R is the known resistance S is the unknown resistance R AC = ρ L AC A = R 0 x L = R 0 x L . . . . . . . . . . ( 2 ) Similarly, R BC = ρ L BC A = R 0 100 - x L = R 0 100 - x L . . . . . . . 3 Now, substituting the equation (2) and (3) in equation (1) we get, R R 0 x L = S R 0 100 - x L R x = S 100 - x S = R 100 - x x This is the required equation to find the value of unknown resistance from the known resistance.

Write the comparison between resistance and specific resistance of a conductor.

Resistance 1. The abstractions offer by the conductor to the flow of current is called resistance. 2. It depends on length and area of cross of the conductor. 3. Its unit is ohm (Ω) Specific resistance 1. Specific resistance (P) is the resistance of material of wire of length 1 m and area of cross-section 1 m 2 2. It does not depend on length and area of cross-section of conductor. 3. Its unit is ohm × meter.

A network of 4 ohm, 4 ohm and 3 ohm resistance are connected to a 10 volt battery as shown in figure. Compute the equivalent resistance and current obtained from battery.

Given, Resistance R 1 = 4 Ω Resistance R 2 = 4 Ω Resistance R 3 = 3 Ω Electromotive force emf = 10 V R 1 and R 2 are in parallel order Now, 1 R P = 1 R 1 + 1 R 2 1 R P = 1 4 + 1 4 1 R P = 1 + 1 4 = 2 4 R P = 2 Ω R P and R 3 are in series order Now, R S = R P + R 3 R S = 2 + 3 R S = 5 Ω Therefore, current resistance in the circuit is 5 Ω . Battery current I = E R E = 10 V R = 5 Ω I = 10 5 I = 2 Ampere Therefore, current of battery is 2 Ampere . Hence, it is the required answer.

Resistance of a resistance thermometer at 0 C o is 6 Ω . At 100 C o is 7 Ω and at unknown temperature is 8 Ω . Find out unknown temperature.

Let R ' be the resistance at any temperature (in Kelvin), R ° be the resistance at 0 ° C or 273 . 15 K ∆ T be the temperature difference from 0 ° C and K be temperature coefficient of resistance change. We have R ° = 6 ohms R ' = R ° ( 1 + K ∆ T ) At 100 ° C or 373 . 15 Kelvin 7 = 6 ( 1 + K ( 100 ) ) 100 K = 7 6 - 1 K = 1 600 per Kelvin For temperature at 8 ohms 8 = 6 1 + 1 600 ∆ T 1 600 ∆ T = 8 6 - 1 ∆ T = 2 6 x 600 ∆ T = 200 Kelvin T - 273 . 15 = 200 T = 473 . 15 Kelvin = 200 ° C Hence, it is the required answer.