__Wheatstone bridge and Potentiometer__

Wheatstone bridge is an arrangement of four resistances which can be used to measure one of them in terms of rest. Here arms AB and BC are called ratio arm and arms AC and BD are called conjugate arms.

**Balanced Wheatstone bridge:**

The bridge is said to be balanced when deflection in galvanometer is zero i.e. no current flows through the galvanometer or in other words VB = VD. In the balanced condition P/Q = R/S, on mutually changing the position of cell and galvanometer this condition will not change.

If the bridge is not balanced current will flow from D to B if VD > VB i.e. (VA – VD) < (VA – VB) which gives PS > RQ.

⇒ S' = 4×11/9 = 44/9 ⇒ 1/S' = 1/r + 1/6

⇒ 9/44 – 1/6 = 1/r ⇒ r = 132/5 = 26.4 Ω

Solved example 2: A voltmeter having a resistance of 998 ohms is connected to a cell of emf 2 volt and internal resistance 2 ohm. The error in the measurement of emf will be

(A) 4 ×10–1 volt (B) 2 ×10–3 volt

(C) 4 ×10–3 volt (D) 2 ×10–1 volt

Solution: (C) Error in measurement = Actual value – Measured value

Actual value = 2A

i = 2/998+2 = 1/500 A

Since E = V + ir = ⇒ V = E – ir = 2 – 1/500 × 2 = 998/500 V

Measured value = 998/500 V ⇒ Error = 2 – 998/500 = 4 × 10–3 volt.

J = sliding point, K = Key, R = Resistance of potentiometer wire, r = Specific resistance of potentiometer wire. Rh = Variable resistance which controls the current through the wire AB

(i) The specific resistance (r) of potentiometer wire must be high but its temperature coefficient of resistance (a) must be low.

(ii) All higher potential points (terminals) of primary and secondary circuits must be connected together at point A and all lower potential points must be connected to point B or jockey.

(iii) The value of known potential difference must be greater than the value of unknown potential difference to be measured.

(iv) The potential gradient must remain constant. For this the current in the primary circuit must remain constant and the jockey must not be slided in contact with the wire.

(v) The diameter of potentiometer wire must be uniform everywhere.

__METER BRIDGE__
In case of Meter Bridge, the resistance wire AC is 100 cm long. Varying the position of tapping point B, bridge is balanced. If in balanced position of bridge AB = l, BC = (100 – l)

So that .Q/P = (100–l)/l

Also P/Q = R/S

S = (100–l)/l R

So that .Q/P = (100–l)/l

Also P/Q = R/S ⇒ S = (100–l)/l R

Solved example 1: In Wheatstone bridge P = 9 ohm, Q = 11 ohm, R = 4 ohm and S = 6 ohm. How much resistance must be put in parallel to the resistance S to balance the bridge

(A) 24 ohm (B) 44/9 ohm (C) 26.4 ohm (D) 18.7 ohm

**Potentiometer**

Potentiometer is a device mainly used :

to measure emf of a given cell

to compare emf's of cells

to measure internal resistance of a given cell

Circuit diagram: Potentiometer consists of a primary circuit and the secondary circuit primary circuit

1. a long resistive wire AB of length L (about 6 m to 10 m long) made up of mangnine or constantan

2. a battery of known voltage e and internal resistance r called supplier battery or driver cell. Secondary circuit

1. One terminal of another cell (whose emf E is to be measured) is connected at one end of the main circuit and

2. the other terminal at any point on the resistive wire through a galvanometer G. This forms the. Other details are as follows

If V > E then current will flow in galvanometer circuit in one direction

If V < E then current will flow in galvanometer circuit in opposite direction

If V = E then no current will flow in galvanometer circuit this condition to known as null deflection position, length l is known as balancing length.

In balanced condition E = xl

or E = xl = V/L l = iR/L l = (e/R+Rh+r) × R/L × l

If V is constant then L ∝ l ⇒ x1/x2 = L1L2 = l1/l2

Standardization of Potentiometer: The process of determining potential gradient experimentally is known as standardization of potentiometer.

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