## To use a meter bridge the resistance of the material of a given wire.

How to Use a Meter Bridge The Resistance of the Material of a Given Wire. |

**A) APPARATUS REQUIRED**

i) MeterBridge

ii) Four Way Key(Commutator)

iii) Resistance Box

iv) Leclanche Cell

v) Connecting Wires

**B) Theory:-** The principle used in the determination of resistance by meter bridge lies in the application of the Wheatstone bridge.

According to wheatstone bridge, principle of ‘P’, ‘Q’, ‘R’, and ‘X’ be four resistance where ‘X’ is unknown resistance that needs to be determined. ‘G’, a moving will galanometer and ‘E’, a leclanche cell, then null deflection in the galannometer

$$\frac PQ=\frac XR\;$$ ——equation(i)

Now, if the above wheatstone bridge circuit be applied in the meter bridge, the circuit will be as shown in the diagram,

Let ‘B’ be the point along the bridge wire when the galvanometer deflection is zero. At balance point, if AB is ‘L’ cm, then BC=(100-L).

Now from equation (i), we have,

$$\frac PQ=\frac XR\;$$

Figure———————–

$$\frac L{(100-L)}=\frac XR$$

$$\therefore\;P\propto AB$$ i.e. $$P\propto L$$

and

$$Q\propto BC$$ i.e. $$Q\propto(100-L)$$

$$\therefore\;\frac PQ=\frac L{(100-L)}$$

$$or,\;X=\left(\frac L{100-L}\right)\;\times\;R\;\;\;\;\;\;\;\;–\;equation\;(ii)$$

If $$’\rho’$$ is the resistivity of the material of the wire, ‘A’ is the area of cross section of wire, and ‘L’ is the length of the wire used then,

$$\rho\;\;=\;\frac{XA}L$$

Since, $$A\;=\;\frac{\mathrm{πd}^2}4$$, where ‘d’ is the diameter of the wire.

∴$$\rho\;=\;\frac{\mathrm{πd}^2\mathrm X}{4L}$$

**C) OBSERVATION**

Length of the given wire (L) = 59cm = 0.59m

Diameter of the given wire (d) = 0.5mm = $$0.5\;\times\;10^{-3}m$$

OBSERVATION TABLE

No. of obs | R(ohm) | l(m) | $$x=\frac l{100-l}\times R$$ | Mean ‘X’ ohm |

1. | 2 | 16.5 | $$X_1=0.3952$$ | $$X_m=1.47965$$ |

2. | 10 | 2.5 | $$X_2=2.5641$$ |

Therefore the Mean value of ‘X’=1.47965

**D) CALCULATION**

$$X_{mean}=\frac{X_1+X_2}2\;=\;\frac{0.3952\;+\;2.5641}2\;=\;1.47965$$

**E) RESULT**

Hence, the resistance of used wire is 1.47965

**F) PRECAUTION**

i) The given resistance wire should be straight and the electrical connections must be tight.

ii) The current should not be passed continuously for a very long time.

iii) The diameter of the wire should be measured very carefully.

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