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# ELECTRIC MEASUREMENT

## RESISTIVITY AND CONDUCTIVITY

The resistance of a wire maintain at a constant temperature is related to its length L and its cross-sectional area (A) by the expression

R = ρl

A

Where ρ is a constant known as resistivity of the material (its unit is ohm-metre, Ωm)

ρ = RA

l

R = resistance, A = cross-sectional area, l = length of the wire.

The resistance is the ability of a material to oppose the flow of current through it.  The greater the resistivity of a wire the poorer it is as an electrical conductor. That is why conductivity is used to specify the current –carrying ability of a material.  The greater the conductivity of a material, the more easily can current flow through the material.  Hence, materials with high conductivity will have low resistively.

Conductivity, σ is the reciprocal of the resistivity

σ = 1

ρ

Electrical Conductivity:  This is a measure of the extent to which a material will allow current to flow easily through it when a p.d is applied at a specified temperature.  It is the reciprocal of the resistivity.

## GALVANOMETER CONVERSION

Conversion of galvanometer to ammeter (Shunt)

An ammeter is used for measuring currents.  A galvanometer is used for detecting and measuring very small currents.  We can convert galvanometer into ammeter by connecting a suitable resistor in parallel with the galvanometer, this is known as shunt. A shunt is a low resistance wire and is used to divert a large part of the current being measured but to allow only a small current to pass through the galvanometer

Conversion of Galvanometer to Voltmeter

A galvanometer used for measuring very small current can be converted to voltmeter by connecting a high resistance or multiplier in series with the galvanometer.

## ONLINE WORK

1. A steady current of 1.5A flows through a copper wire of length 10m and cross-sectional 3.44×10-8m2. What is the voltage applied across the wire if the resistivity of copper is 1.72 x10-8Ωm?
2. A galvanometer of resistance 5 ohms gives a full scale deflection when a current of 50mA flows through it. How will you convert it to an ammeter capable of measuring 2A?

## ASSIGNMENT

SECTION A

1. When a resistance r is a across a cell, the voltage across the terminals of the cell is reduced to two-thirds of its nominal value. The internal resistance of the cell is (a) 1/3R (b) ½R (c) 2/3R (d) R
2. Which of the following does not determine the electrical resistance of a wire? (a) Length (b) Mass (c) Cross-sectional area (d) Temperature
3. A wire of 5Ω resistance is drawn out so that its new length is twice the original length. If the resistivity of the wire remains the same and the cross-sectional area is halved, the new resistance is (a) 5Ω (b) 10Ω (c) 20Ω (d) 40Ω
4. A cell of e.m.f 1.5V and internal resistance 2.5ohms is connected in series with an ammeter of resistance 0.5 ohms and a resistor of resistance 0.7ohms. Calculate the current in the circuit (a) 6.67A (b) 0.20A (c) 0.60A (d) 0.15A
5. A cell of internal resistance 2 ohms supplies a current of a 6-hm resistor. The efficiency of the cell is (a) 12.0% (b) 25.0%      (c) 33.3% (d) 75.0%

SECTION B

1. A galvanometer of resistance 50Ω which gives a full-scale deflection for 1mA is to be adapted to measure currents up to 5A. (i) calculate the resistance of the resistor required (ii) if the resistor is made of a material of cross-sectional area 4×10-4cm2 and resistivity 2×10-6Ωm, calculate its length
2. A battery of three cells in series, each of e.m.f. 2 V and internal resistance 0.5 Ω is connected to a 2 Ω resistor in series with a parallel combination of two 3Ω resistors. Draw the circuit diagram and calculate (i) the effective external resistance (ii) the current in the circuit (iii) the lost volts in the battery (iv) the current in one of the 3 Ω

Calculate the length of a constantan wire of diameter 0.6 mm and resistivity 1.1 x 10-6 Ωm required to construct a standard resistor of resistance 35Ω

See also

ELECTROLYSIS

ELECTRIC CELLS

ELECTRIC FIELD

GRAVITATIONAL FIELD

ELECTROMAGNETIC WAVES

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