Classwork Series and Exercises {Physics- SS3}: Electric Measurement

Resistivity

Resistivity: The resistance of a wire or material conductor maintained at a constant temperature is related to its length (l) and its cross-sectional area (A) by the expression

R = ρ l/A

Where ρ is a constant of proportionality known as the resistivity of the material

Then we could deduce ρ = RA/l

Thus ρ = R when l = 1 and A = 1 Hence we define resistivity as follows:

Resistivity is the resistance of unit length of material of unit cross-sectional area.

Where R is measured in ohms, A in m, l in m, the unit of ρ is in ohm-meter (Ωm).

We recall our definition of resistance as 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. Because of this, the term conductivity is used to specify the current-carrying ability of a material. The greater the conductivity the more easily current flows through the material. Thus materials of high conductivity also have low resistivity. Conductivity, σ =l/ρ

Electrical conductivity 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 specific temperature. It is the reciprocal of resistivity.

Example: The resistance of a wire of length 100 cm and diameter 0.3 mm is found to be 3.0 ohms. Calculate (i) the resistivity, (ii) the conductivity of the material of the wire.

Solution

(i) Resistivity

ρ = RA/l

R= 3 Ω, l = 100 cm = 1.0 m, r = (0.3/2 x 10^-3) m

Area = πr² = π(0.3/2 x 10^-3)² m²

         = π(1.5 x 10^-4)² m²

ρ = 3 x 22//7 x (1.5 x 10^-4)²/1 Ωm

  = 21.21 x 10^-8 Ωm

  = 2.12 x 10^-7 Ωm

(ii) Conductivity, σ = 1/ρ

                           = 1/2.12 x10⁼⁷ (Ωm)^-1

                           = 4.7 x 10⁶ (Ωm)^-1

Conversion of Galvanometer to Ammeter

An ammeter is used for measuring electric currents. A galvanometer is used for detecting and measuring very small currents.

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

Since Galvanometer is a very sensitive instrument therefore it can’t measure heavy currents. In order to convert a galvanometer into an Ammeter, a very low resistance known as “shunt” resistance is connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes through the shunt. In this way a Galvanometer is converted into Ammeter and can measure heavy currents without fully deflected

An ammeter is always connected in series to a circuit

Then current through shunt:

Is = (I-Ig)

potential difference across the shunt:

Vs= IsRs

Vs = (I – Ig)Rs ——-(ii)

But      Vs =Vg (I – Ig)Rs = IgRg

Rs = Ig Rg/I – Ig

Voltmeter is an electrical measuring device, which is used to measure potential difference between two points in a circuit.

Connection of Voltmeter in Circuit Voltmeter is always connected in parallel to a circuit.

 If potential between the points to be measured = V and if galvanometer gives full-scale deflection, when current “I_g” passes through it. Then,

V = I_g (R_g + R_x)
V = I_gR_g + I_gR_x
V – I_gR_g = I_gR_x
R_x = (V – I_gR_g)/I_g
R_x = V/I_g – R_g

Measurement of Resistance by Ammeter-Voltmeter Method

This is very popular method for measurement of medium resistances since instruments required for this method are usually available in laboratory. The two types of connections employed for ammeter voltmeter method are shown in figures below. In both the methods if readings of ammeter and voltmeter are taken then we can measure value of resistance by using formula:

R_m = voltmeter reading/ammeter reading = V/I

The measured value of resistance R_m, would be equal to the true value R, if the ammeter resistance is zero and the voltmeter resistance is infinite, so that the conditions in the circuit are not disturbed. But in actual practice this is not possible and hence both methods give inaccurate results.

Consider circuit of figure (a):

Voltmeter-Ammeter method

In this method ammeter measures the true value of current flowing through resistance but voltmeter does not measures the true value of the voltage across the resistance. The voltmeter indicates the sum of the voltage across resistance and ammeter.

Let Ra be the resistance of the ammeter.

Therefore, voltage across the ammeter

V_a = IR_a

Measured value of the resistance

R_m1 =V/I = V_{r +} V_a/I = I_r + IR_a/I = R + R_a

Therefore, true value of resistance, R = R_m1 – R_a

= R_m1(1 – R_a/R_m1)

Hence the measured value of the resistance is higher than the true value. It is also clear from the above equation that the true power is equal to the measured only if the ammeter resistance is zero.

Relative error, E_r = R_m1 – R/R = R_a/R

It is clear from the above equation that the error will be small if the value of the measuring resistance is large as compare to the internal resistance of the ammeter .therefore circuit should be used when measuring resistances are high.

Consider circuit of figure (b):

Voltmeter-Ammeter method

In this circuit the voltmeter measures the true value of the voltage across the measuring resistance but the ammeter does not measure the true value of the current flowing through the resistance. The current through the ammeter is the sum of the current through the voltmeter and resistance.

Let R_v be the resistance of the voltmeter.

Therefore current through the voltmeter, I_v = V/R_v

Measured value of the resistance R_m2 = V/I = V/I_{r +} I_v = V/V/R +V/R_v = R/1 +R/R_v

True value of resistance;

R = R_m2R_v/R_v – R_m2 = R_m2 (1/1 – R_m2/R_v)

From the above equation it is clear the true value of the resistance will be equal to the measured value only when the voltmeter resistance is equal to the infinite. However, if the resistance of the voltmeter is very large as compared to the resistance under measurement:

R_v >> R_m2

And therefore R_m2/R_v is very small

We have  R = R_m2 (1 + R_m2/R_v)

Thus, the measured value of the resistance is smaller than the true value.

Relative error, E_r = R_m2 – R/R = R²_m2/R_v R

The value of R_m2 is approximately equal to E_r = -R/R_v

It is clear from the above equation that the relative error will be small if the resistance under measurement is very small as compared to the resistance of the voltmeter .hence the circuit should be used when the measuring values of resistances are low.

The Wheatstone Bridge

A Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. Its operation is similar to the original potentiometer. It was invented by Samuel Hunter Christie in 1833 and improved and popularized by Sir Charles Wheatstone in 1843. One of the Wheatstone bridge’s initial uses was for the purpose of soils analysis and comparison

 

In the figure, is the unknown resistance to be measured; R₁, and are resistors of known resistance and the resistance of is adjustable. If the ratio of the two resistances in the known leg (R₂/R₁) is equal to the ratio of the two in the unknown leg (R_x/R₃), then the voltage between the two midpoints (B and D) will be zero and no current will flow through the galvanometer V_g. If the bridge is unbalanced, the direction of the current indicates whether R₂ is too high or too low. R₂ is varied until there is no current through the galvanometer, which then reads zero.

Detecting zero current with a galvanometer can be done to extremely high accuracy. Therefore, if R₁, R₂ and R₃ are known to high precision, then R_x can be measured to high precision. Very small changes in R_x disrupt the balance and are readily detected.

At the point of balance, the ratio of

R₂/R₁ = R_x/R₃

Þ  R_x = R₂/R_{1 .} R₃

Alternatively, if R₁, R₂ and R₃ are known, but R₂ is not adjustable, the voltage difference across or current flow through the meter can be used to calculate the value of R_x, using Kirchhoff’s circuit laws (also known as Kirchhoff’s rules). This setup is frequently used in strain gauge and resistance thermometer measurements, as it is usually faster to read a voltage level off a meter than to adjust a resistance to zero the voltage.

Derivation

First, Kirchhoff’s first rule is used to find the currents in junctions B and D:

I₃ – I_x + I_G = 0

I₁ – I₂ – I_G = 0

Then, Kirchhoff’s second rule is used for finding the voltage in the loops ABD and BCD:

(I₃ . R₃) – (I_{G .}R_G) – (I₁ . R₁) = 0

(I_x . R_x) – (I₂ . R₂) + (I_G . R_G) = 0

When the bridge is balanced, then I_G = 0, so the second set of equations can be rewritten as:

I_{3 .} R₃ = I₁ . R₁

I_x . R_x = I₂ . R₂

Then, the equations are divided and rearranged, giving:

R_x  = I₂ . R₂ . I_{3 .} R₃/ I₁ . R₁. I_x

From the first rule, I₃ = I_x and I₁ = I₂. The desired value of R_x is now known to be given as:

R_x = R₃ . R₂/R₁

If all four resistor values and the supply voltage (V_S) are known, and the resistance of the galvanometer is high enough that I_G is negligible, the voltage across the bridge (V_G) can be found by working out the voltage from each potential divider and subtracting one from the other. The equation for this is:

V_G = (R_x/R₃ + R_x – R₂/R₁ + R₂) V_s

where V_G is the voltage of node B relative to node D.

Significance

The Wheatstone bridge illustrates the concept of a difference measurement, which can be extremely accurate. Variations on the Wheatstone bridge can be used to measure capacitance, inductance, impedance and other quantities, such as the amount of combustible gases in a sample, with an explosimeter. The Kelvin bridge was specially adapted from the Wheatstone bridge for measuring very low resistances. In many cases, the significance of measuring the unknown resistance is related to measuring the impact of some physical phenomenon (such as force, temperature, pressure, etc.) which thereby allows the use of Wheatstone bridge in measuring those elements indirectly.

EXERCISES

Lets see how much you’ve learnt, attach the following answers to the comment below

1. If a wire has a resistance of 1.32Ω, a length of 110 cm and a n area of cross-sectional of 0.00415 cm², Find the resistivity of the material of which it is made. A. 6.45 x 10^-7Ωm B. 4.98 x 10^-7Ωm C. 3.46 x 10^-7Ωm D. 2.39 x 10^-7Ωm
2. A galvanometer has as resistance of 5Ω. By using a shunt wire of resistance 0.05Ω, the galvanometer could be converted to an ammeter, capable of reading 2A. What is the current through the galvanometer? A. 0.02A B. 0.06A C. 2.09A D. 4.15 A
3. A Wheatstone bridge is an electrical circuit used to measure an unknown ……………… by balancing two legs of a bridge circuit, one leg of which includes the unknown component. A. Potential difference B. electrical resistance C. Voltage D. Current
4. In other to concert a galvanometer into ammeter, the shunt resistance must be connected in …………… to the galvanometer. A. Series B. Parallel C. direct  D. Indirect
5. A resistance wire of length 2m and of uniform cross-sectional area 5.0 x 10^-7 m² has a resistance of 2Ω. Calculate its resistivity. A. 5.0 x 10^-6 m B. 5.6 x 10^-7 m² C. 5.0 x 10^-7 m²   D. 4.7 x 10^-5 m²

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