Electrical

# Ohm’s Law: Statement, Formula, Solved Examples

## Ohm’s Law

Ohm’s Law: Statement, Formula, Solved Examples- In any electric circuit, the current tends to be directly proportional to an applied voltage and inversely proportional to the total resistance of this circuit, provided that no change occurs in the physical condition of the circuit (i.e., circuit’s temperature remains constant). This is called ohm’s law, which has been presented by Dr. John Ohm.

“In any electric circuit, the current is directly proportional to the applied voltage and inversely proportional to the total circuit resistance”

The aforementioned description can be written in the form of an equation as follows;

Current = Potential difference / Resistance

Or, I = V / R

Or, Effect = Cause / Opposition

This equation is generally called ohm’s law equation. The equation of this law can be applied on a whole circuit or a part of this circuit as well. When we apply this law on an entire circuit, then total circuit current is obtained by dividing the total voltage by total circuit resistance. If we are required to determine a current flowing through a specific part of the circuit, then just voltage of that part is divided by the resistance of that particular part.

For solving arithmetic calculations, ohm law can be used in the following three forms;

I = V / R or V = IR or R = V / I

## Laws of Resistance

That characteristic of any object or substance, due to which it creates obstacles or causes opposition to the flow of electrons or current from within it, is called resistance.

The resistance of a conductor depends on the following factors, which are called laws of resistance.

1. The resistance of a conductor is directly proportional to its length (l). That’s, resistance of a conductor increases as a result of an increase in its length. And its resistance decreases owing to a decrease in its length.
2. The resistance of a conductor is inversely proportional to its cross-sectional area, which means that resistance of a thick wire tends to be low, whereas the resistance of a thin wire tends to be high.

III. The resistance of a conductor also depends on the nature of conductor material. For example, the resistance of two similar conductors having equal length and thickness, made from different kind of materials, tend to be different.

1. Resistance also depends on the temperature of a conductor. That’s resistance changes as a result of variations in temperature.

If the last factor is ignored for the time being, then the following relationship develops amongst the aforementioned factors.

R α ℓ and R α 1 / a or R α ℓ / a or R =  / a

Here,  (rho) is a constant, which depends on the nature of a conductor material and it is known as a specific resistance or resistivity of the conductor. If values of both a and ℓ becomes same, then value of  becomes equal to R.

The specific resistance of any material actually tends to be a resistance of its unit length and unit cross-section. In other words, it can also be described as below;

The resistance found between two opposite surfaces of any piece of material having a size of one meter cube (or one centimeter cube), is called specific resistance. This has been illustrated in figure 1.9 below;

It refers to a resistance found between opposite faces of a meter (or centimeter) cube (or unit cube of that material. The unit of specific resistivity is ohm- meter. However, if length of a cube is termed in centimeters and area (A) in square centimeters, then rho unit tends to be ohm – cm (i.e., a conductor having a length equal to 1 cm and its cross – sectional area 1 cm2, then resistance of such a conductor will be termed as resistivity.

 Type of Material Materials Resistivity (ohm – cm) Insulators Glass Mica Rubber Quartz 17.0 x 1012 1.0 x 1012 1.0 x 1018 1.0 x 1019 Semiconductors Carbon Silicon Germanium 1.0 x 10-3 2.0 x 105 65.0 Conduct x 4 /10 4ors Silver Copper Aluminum Tungsten Brass Iron Mercury 1.6 x 10-6 1.7 x 10-6 2.8 x 10 -6 5.6 x 10-6 6.0 x 10-6 10.5 x 10-6 9.6 x 10-5

## Calculations about Laws of Resistance

Example; A length of wire has a resistance of 4.5 ohms. Find the resistance of another wire of the same material, three times as long and twice the cross – sectional area.

Solution;

Resistance of wire = 4.5 x 3 / 2 = 6.75 ohms

Example 2; Find the resistance of 1000 yards of copper wire 0.02 in in diameter. Take the specific resistance of copper as 2/3 microohm – in.

Solution;

R = ρ l / A l = 1.000 x 36 in …  {1 yard = 3 ft. and 1 ft. = 12 inch, i.e., 1 yard = 36 inch}

Ρ = 2 / 3 x 10 5 ohm

A = π / 4 d2 = π /4 x [2 / 100]2 sq.in = π / 4 x 4 / 104 = π/ 104 sq.in.

R = 2 x 1000 x 36 x 10 4 / 3 x 10 6 x 3.1416 = 72 x 10 7 / 9.4248 x 10 6 = 76.39 ohms

Example 3; If 5 yards of manganin wire 0.05 inch in diameter has a resistance of 1.7 ohms, find the specific resistance of the material.

Solution;

R = ρ I l A

A = π / 4 x (5 / 100)2 = π / 4 x 25 / 10000

ρ = RA/ l

A = π / 4 x 1 / 400 = π / 1600 sq. inch

ρ = 1.7 x π / 1600 x 5 x 36 = 0.00001854 ohm – inch = 18.54 microohm – inch

Example 4; The resistance of 500 meter of a certain wire is 125 ohms. What length of the same wire will have a resistance of 90 ohm?

Solution;

Here resistance R1 be 125 ohms, having length 500 meter when resistance is 90 Ω. Then

R1 =  /1 and R2  l2

Or R1 / R2 = l1 / l2

l2 = R2 l1 / R1 = 90 x 500 / 125 = 360-meter, Ans.

Example 5; If the specific resistance of manganin is 41.8  Ω cm. What is the diameter of a manganin wire 68.7 meter long which has a resistance of 8.86 ohms?

Solution;

The resistance is 8.86 ohms, the specific resistance 41.8  Ω cm. and length 68.7 meter.

Now area of cross section = π d2 / 4 = ρ. l / R

And d2 = 4 ρ. l / π R

= 4 x 41.8 x 10 -6 x 68.7 x 100/ 3.14 x 8.86

And d2 = 0.04128 and d = √0.04128 = 0.203 cm. Ans

Example 6; A coil consists of 2000 turns of copper wire having a cross – sectional area of 0.8 mm2. The mean length per turn is 80 cm. and the resistivity of copper is 0.02  Ω-m. Find the resistance of the coil and power absorbed by the coil when connected across 110 V DC supply?

Solution;

Length of the coil, l = 0.8 x 2000 = 1600 m; A = 0.8 mm2 = 0.8 x 10 -6 m2

R = ρ l / A = 0.02 x 10 -6 x 1600 / 0.8 x 16 -6

Power absorbed = V2 / R = 110 2 / 40 = 302.5 W

Example 7; A rectangular carbon block has dimensions 1.0 cm x 1.0 cm x 50 cm. (i). What is the resistance measured between the two square ends? (ii). Between two opposing rectangular faces? Resistivity of copper may be taken as 1.72 x 10 -8 Ω-m.

Solution;

(i). R = ρ // A

Here, A = 1 x 1 = 1 cm2 = 10-4 m2; I = 0.5 m

R = 3.5 x 10-5 x 0.5/ 10-4 = 0.175 Ω

(ii). Here, I = 1 cm; A = 1 x 50 = 50 cm2= 5 x 10-3 m2

R = 3.5 x 10-5 x 10-2 / 5 x 10-3 = 7 x 10-5

Example 8; Calculate the resistance of 1 km long cable composed o 19 strands of similar copper conductors, each strand being 1.32 mm in diameter. Allow 5% increase in length for “lay” twist of each strand in completed cable. Resistivity of copper may be taken as 1.72 x 10-8 Ω-m.

Solution;

Allowing for twist, the length of the strands = 1000 m + 5% of 1000 m = 1050 m

Area of cross – section of 19 strands of copper conductors is

19 x π x d2 / 4 = 19 π x (1.32 x 10 -3)2 / 4 m2

Now, R = ρ l / A = 1.72 x 10-8 x 1050 x 4 / 19 π x 1.32 2 x 10 -6 = 0.694 Ω

Example 9; The resistance of the wire used for telephone line is 35 Ω per kilometer when the weight of the wire is 5 kg per kilometer. If the specific resistance of the material is 1.95 x 10 -8 Ω – m, what is the cross – sectional area of the wire? What will be the resistance of a loop to a subscriber 8 km from the exchange if wire of the same material but weighing 20 kg per kilometer is used?

Solution;

Here R = 35 Ω; l = 1 km = 1000 m; ρ = 1.95 x 10 -8 Ω – m

Now, R = ρ l / A or A = ρl / R ∴ A = 1.95 x 10 -8 x 1000 / 35 = 55.7 x 10 -8 m2

In the second case, if the wire is of the same material but weighs 20 kg / km, then its cross – section must be greater than that in the first case.

Cross – section in the second case = 20 / 5 x 55.7 x 10 -8 = 222.8 x 10 -8 m2

Length of wire = 2 x 8 = 16 km = 16000 m ∴ R = ρ l / A = 1.95 x 10 -8 x 16000 / 222.8 x 10-8

= 140 Ω Ans.

Example 10; Determine the resistance of 91.4 meters annealed copper wire having a cross – section of 1.071 sq. cm, resistance of copper being 1.724 microohms. Cm at 20℃.

Solution;

Here, l = 91.4 meters = 9140 cm

A = 1.071 sq. cm

And ρ = 0.000001724 ohm – cm.

Now, resistance R = ρ l / A

Substituting various values, we have

R = 0.000001724 x 9140 / 1.071 = 0.0145 ohms. Ans

Example 11; Find the area of cross – section and diameter of a wire of copper to have a resistance of 0.13 ohm per km.

Given ρ for a copper = 1.7 x 10 -6 Ω per cubic centimeter

Solution;

Since R = ρ l / A ∴ A = ρ x l / R

Taking ρ = 1.7 x 10 -6 Ω / cm3 and l = 1 km = 1000 m = 105 cm,

We have A = 1.7 x 10 -6 x 10 5 / 0.13 = 1.31 sq. cm

Now, cross – sectional area, A = π d2 / 4; where d is the diameter;

d = √4A / π = √ 4 x 1.31 / π = 1.29 cm. Ans

Example 12; Calculate the resistance of a wire 20 m long and 2 mm radius having resistivity 5.5 x 10 -8 ohm-m. if the two ends of this wire are connected to a 24-volt source, what is the value of current?

Solution;

Length of wire “L” = 20 m

Radius of wire “r” = 2 mm = 2 x 10 -3 m

Area of cross – section “A” = π r2 = 22/ 7 x (2 x 10-3)2 = 12.6 x 10 -6 m2

Resistivity “ρ” = 5.5 x 10 -8 ohm -m

Resistance “R” = ρ L / -8 x 20 / 12.6 x 10 -6 = 9 A = 5.5 x 10 -8 x 20 / 12.6 x 10 -6 = 8.73 x 10 -2 ohm

Current “I” = V / R = 24 / 8.73 x 10 -2 = 274 A

## Effects of Temperature on Resistance

As has been discussed above, the resistance of any material cannot be constant, rather it keeps changing as a result of a change in temperature. The effects which are cast by a change in temperature on the resistances of different materials, are as follows;

(i). The resistance of pure metals tends to increase due to an increase in temperature. This increase in resistance tends to be extremely high and quite regular for normal ranges of temperature. If a graph is drawn between resistance and temperature, it appears in the shape of a straight line. (See figure 1.10)

Figure 1.10 – resistance / Temperature Graph

Remember that if resistance of a conductor continues increasing due to an increase in temperature, we can describe it as such that temperature coefficient of a conductor is positive.

(ii). Alloy’s resistance increases as a result of an increase in temperature, however this increase in resistance tends to be very slow and irregular. The resistance of alloys being used in electrical work, practically have constant resistant nearly at all temperatures. For example, Eureka 940% nickel and 60% copper) and Magainin, etc.

(iii). The resistance of non – metallic items e.g., paper, rubber, glass, mica, and electrolyte etc. Tends to reduce as a result of an increase in temperature. (Electrolyte is a type of solution, through which electricity can transmit quite easily, e.g., water containing sulphuric acid). The resistance of an insulator declines considerably due to an increase in its temperature. In other words, their temperature coefficient tends to be negative.

(iv). The resistance of copper, manganese, and nickel (which are known as thermistors) tends to be relatively high at low temperature. However, their resistances decline very rapidly as a result of an increase in temperature. That’s the reason, they are mostly used for the regulation of voltage in circuits.

## Temperature Co-efficient

An amount of change in the resistance value resulting from a one degree change in temperature, is known as temperature coefficient of that resistance. It is denoted by “𝛼” (alpha).

Temperature co-efficient of resistance is the amount by which resistance changes when temperature changes one degree centigrade.

Suppose that if resistance of a conductor at ℃ tends to be R0 ohm, and if it is heated up to t℃, then its resistance becomes Rt. This increase or change in resistance resulting from an increase or a change in temperature, is denoted by the following equation;

Change in Resistance, Δ R = R t – R 0

Changes taking place in resistance depends on the following factors;

(i). Changes in resistance are directly proportional to its initial resistance.

(ii). It is directly proportional to changes in temperature.

(iii). It depends on the nature of conductor material.

The points mentioned above can also be expressed in the form of an equation as follows;

(1). Rt – R0 𝛼 R x t or Rt – R0 = 𝛼 R0 – t

Here, 𝛼 (alpha) is a constant, and it is known as a temperature coefficient of resistance. The above equation can also be written as follows;

𝛼 = Rt – R0 / R0 x t = Δ R = R t – R 0

If R0 is one ohm and t is one degree centigrade, then

𝛼 = Δ R = R t – R 0

Therefore, temperature coefficient of any material can be defined in the following ways;

The increase in resistance per ohm original resistance per ℃ rise in temperature is called temperature coefficient.  The following relation is elicited from equation (1).

(2). Rt = R0 (1 + 𝛼 t)

It must be kept in mind that this relation sustains even if temperature increases or decreases or even if no change takes place in temperature. As soon as temperature of any conductor declines, its resistance also gets lower.

In figure 1.10, temperature resistance graph of copper has been illustrated, which practically happens to be straight line. If this line is progressed backwards, then this resistance line will intersect the temperature line at that point, where the temperature tends to be -234.5 ℃. It means that copper’s conductor resistance at this point will become zero. The following relation can be represented by two similar triangles shown in the figure.

Rt / R0 = t + 234.5 / 234.5 = [ 1+ t / 234.5]

Rt = R0 [1 + t / 234.5] or Rt = R0 (1 + 𝛼t)

Where 𝛼 = 1 / 234.5 = 0.00428 (for copper)

 Temp. in ℃ 0 5 10 20 30 40 50 𝛼 0.00428 0.00418 0.00409 0.00393 0.00378 0.00364 0.00352

## Calculations About Temperature Co-efficient of Resistance

Example 1; The field winding of a generator has a resistance of 40 ohms at 0℃. What is its resistance at 50℃? Resistance – temperature coefficient of copper is 0.0043 per ℃ at 0℃.

Solution;

Rt = R0 (1 + 𝛼 t)

=40 (1 + 0.0043 x 50)

= 40 x 1.215 = 48.6 ohms

Example 2; A column of mercury has a resistance of 10 ohms at 15℃; what will be its resistance at 30℃? The resistance – temperature coefficient of mercury is 0.0072 per ℃ at 0℃.

Solution;

Rt = R0 (1 + 𝛼 t)

R15 =R0 (1 + 15 𝛼) … (1)

R30 = R0 (1 + 30 𝛼) … (2)

Divide equation (2) by equation (1)

R30 / R15 = R0 (1+30 𝛼) / R0 (1 + 15 𝛼)

i.e., R30 = R15 [1+30 𝛼/ 1 + 15 𝛼]

= 10 x 1+30 x 0.0072 / 1+ 15 x 0.0072

= 10 x 1.216 / 1.108 = 10.97 ohms

Example 3; A coil has a resistance of 10 ohms at 20℃. What is its resistance at 60℃? Resistance – temperature coefficient at 20℃= 0.0039 per ℃.

Solution;

Rt = Rt {1 + 𝛼 (T – t)}

R60 = 10 {1+ 0.0039 (60° – 20°0}

= 10 {1+ 0.0039 x 40}

=10 x 1.156

= 11.56 ohms

Example 4; The shunt winding of a DC generator has a resistance of 55 ohms at 20°C. After a run at full load the resistance rise to 65 ohms. Find the average temperature of the winding. Resistance – temperature coefficient for copper, 0.004 per °C at 0°C.

Solution;

Rt = R0 (1 + 𝛼 t)

65 = R0 (1 + 0.004t) … (1)

55 = R0 (1+ 0.004 x 20) … (2)

i.e., 55 = 1.08 R0

Divide equation (2) by equation (1)

55/65 = 1.08 R0 / R0 (1 + 0.004 t)

55 (1+ 0.044 t) = 65 x 1.08

55 + 0.22 t = 70.2 or 0.22 t = 15.2

t = 55.02 / 0.22 = 69℃

Example 5; A shunt winding of a motor has a resistance of 80 ohm at 15℃. Find the resistance at 50℃. Resistance – temperature coefficient of copper 0.004 per degree at 0℃.

Solution;

R0 = resistance at 0℃, Rt1, resistance at 15℃, Rt2 = resistance at 50℃, 𝛼 the temperature coefficient at 15℃.

Here, Rt1 = R0 (1 + 𝛼t1)

Rt2 = R0 (1 + 𝛼t2)

And Rt2 / Rt1 = [1+ 𝛼t2 / 1 + 𝛼t1]

Rt2 = Rt1 [1 + 𝛼t2 / 1 + 𝛼t1]

=80 [1+ 0.004 x 50 / 1+ 0.004 x 15]

=80 [1 + 0.02 / 1+ 0.06]

=80 x 1.2 / 1.06 = 90.57-ohm Ans

Example 6; A copper coil has a resistance of 0.4 Ω at 12℃. Find its resistance at 52℃. Resistance – temperature coefficient of copper is 0.004 per ℃ at 0℃.

Solution;

The resistance is 0.4 Ω at 12℃ and we know that,

𝛼 = Rt2 – Rt1 / Rt1 (t2 – t1)

Rt2 – Rt1 = Rt1 𝛼 (t2 – t1) or Rt2 = Rt1 + Rt1 𝛼 (t2 – t1)

Or Rt2 =Rt1 [1+ 𝛼 (t2 – t1)]

=0.4 [1 + 0.004 (52 – 12)

= 0.4 [1 + 0.004 x 40]

Rt2 = 0.4612 ohm. Ans

Example 7; The resistance temperature coefficient of phosphor bronze 0.00394 per ℃ at 0℃. A length of phosphor bronze wire has a resistance of 10 ohm at 20℃. What is its resistance at 60℃?

Solution;

The resistance is Rt2 and Rt1 at two different temperatures i.e.,

Rt2 = Rt0 (1 + 𝛼 t2)

Rt1 = Rt0 (1 + 𝛼 t1)

Dividing we have,

Rt2 / Rt1 = 1 + 𝛼 t2 / 1 + 𝛼 t1

Or Rt2 = Rt1 [1 + 𝛼 t2 / 1 + 𝛼 t1]

= 10 [ 1 + 0.00394 x 60 / 1 + 0.00394 x 20]

= 10 [ 1.2364 / 1.0788]

=11.46 ohm. Ans

Example 8; The shunt winding of a motor has a resistance of 45 ohm at 20℃. Find the average temperature at the end of a run when the resistance increased to 48.5 ohms. Resistance temperature coefficient 0.004 per ℃ at0℃.

Solution;

Let Rsh1 be the resistance of shunt winding at 20℃ and Rsh2 after running, i.e., 48.5 Ω.

Rsh2 / Rsh1 = 1 + 𝛼 t2 / 1 + 𝛼 t1

= 48.5 / 45 = 1+ 0.004 t2 / 1 + 0.004 x 20 or 1.078 = 1 + 0.004 t2 / 1.08

1.078 x 1.08 = 1 + 0.004 t2 and t2 = [1.078 x 1.08 – 1 / 0.004]

=0.16424 / 0.004 = 41.06

Final temperature = 41.06℃ Ans.

Example 9; The field coils of a generator have a resistance of 150 ohms at 16℃. After a 6 hour run resistance has increased to 162 ohms. What is the final average temperature of the coils? Resistance temperature coefficient 0.0043 per ℃ at 0℃.

Solution;

The field coils resistance is 150 ohms at 16℃ and temperature coefficient at zero-degree 𝛼 = 0.0043.

R3 / R1 = 1 + 𝛼 t2 / 1 + 𝛼 t1

= 162 / 150 = 1 + 0.0043. t2 / 1 + 0.0043 x 16 or 1.08 = 1 + 0.0043 t2 / 1.0688

1.154304 = 1 + 0.0043 t2 or t2 = 1.154304 – 1 / 0.0043 = 35.88

So, final temperature = 35.88℃ Ans.

Example 10; The resistance of a coil of wire increases from 40 ohm at 10℃ to 48.25 ohms at 50℃. Find the temperature coefficient at 0℃ of the conductor material.

Solution;

We know that R2 / R1 = 1 + 𝛼 t2 / 1 + 𝛼 t1

Or R2 + R2 𝛼 t1 = R1 + R1 𝛼 t2

R2 – R1 = R1 𝛼 t2 – R2 𝛼 t1

R2 – R1 = 𝛼 (R1t2R2t1)

𝛼 = R2 – R1 / R1 t2 – R2t1 = 48.25 – 40 / 40 x 60 – 48.25 x 10

= 8.25 / 2400 – 28.75 = 8.25 / 1017.5

𝛼 = 0.0053 Ans.

Example 11; A coil has a resistance of 20 ohm at 25℃. Find its resistance at 65℃. Resistance temperature coefficient at 25℃ = 0.00385 per ℃.

Solution;

Final resistance = Initial Resistance [1 + 𝛼 (difference of temperature)]

= 20 [1 + 0.00385 (65 – 25)]

= 20 [1 + 0.00385 x 40]

R2 = 23.08 ohms. Ans

Example 12; The field coils of a motor have resistance of 120 ohm at 15℃. After a run at full load, the resistance increases to 135 ohms. Find the average temperature of the coils. Take the resistance temperature coefficient to be 0.00401 per ℃ at 15℃.

Solution;

We know that,

R2 / R1 = 1 + 𝛼 t2 / 1 + 𝛼 t1

135 / 120 = 1 + 0.00401 x t2 / 1 + 0.00401 x 15

1.125 = 1 + 0.00401 x t2 / 1.06015

1.19266 = 1 + 0.00401 x t2

0.00401 x t2 = 1.19266 – 1 = 0.19266

t2 = 0.19266 / 0.00401

= 48℃

Example 13; A platinum coil has a resistance of 3.146 Ω at 40℃ and 3.767 Ω at 100℃. Find the resistance at 0℃and the temperature coefficient of resistance at 40℃.

Solution;

R100 = R0 (1 + 100 𝛼0) … (i)

R40 = R0 (1 + 40 𝛼0) … (ii)

3.767 / 3.146 = 1 + 100 𝛼0 / 1 + 40 𝛼 0

Or 𝛼0 = 0.00379 or 1 / 264 per 0℃

From (i), we have 3.767 = R0 (1 +100 x 0.00379) = R0 = 2.732 Ω

Now,

𝛼40 = 𝛼0 / 1 + 40 𝛼0

= 0.00379 / 1 + 40 x 0.00379 = 1 / 304 per ℃

Example 14; two coils connected in series have resistance of 600 Ω and 300 Ω with temperature coefficient of 0.1% and 0.4% respectively at 20℃. Find the resistance of the combination at a temperature of 50℃. What is the effective temperature coefficient of combination?

Solution;

Resistance of 600 Ω resistor at 50℃ = 600 [1 + 0.001 (50 – 20)] = 618 Ω

Similarly, resistance of 300 Ω resistor at 50℃ is = 300 [1 + 0.004 (50 -20)] = 336 Ω

Hence, total resistance of combination at 50℃ is = 618 + 336 = 954 Ω

Let β = resistance – temperature coefficient at 20℃

Now, combination resistance at 20℃ = 600 + 300 = 900 Ω

Combination resistance at 50℃ = 954 Ω

Final resistance = Initial resistance [1 + (β (t2 – t1)]

954 = 900 [1 + β (50 – 20)]

Β = 0.002, Ans.

Example 15; The temperature coefficient of resistance of phosphorous bronze is 3.94 x 10 -2 Ω per ℃. Find the temperature coefficient of resistance at (a) 20℃ and (b) 100℃.

Solution;

We know that,

𝛼1 = 𝛼0 / 1 + 𝛼0 t1

(a). 𝛼20 = 3.94 x 10-3 / 1 + 3.94 x 10-3 x 20 = 3.94 x 10-3 / 1.0788 = 3.66 x 10-3 Ω/ ℃

(b). 𝛼100 = 3.94 x 10 -3 / 1 + 3.94 x 10 -3 x 100 = 3.94 x 10 -3 / 1.394 = 2.83 x 10 -3 Ω/ ℃

Example 16; At the temperature of 15℃ a p.d of a 250 V is applied to a field coil of a generator; it carries a current of 5A. what will be the temperature of the coil when the current reduces to 4A at the same voltage? Take 𝛼0 for copper as 1 / 2 34.5 Ω/ ℃.

Solution;

R1 = V / I1 = 250 / 5 = 50 Ω

And R2 = V / I2 = 250 / 4 = 62.5 Ω

Since, R1 = R0 (1 + 𝛼0 t1) and R2 = R0 (1 + 𝛼0 t2), on dividing we get

R2 / R1 = 1 + 𝛼0 t2 / 1 + 𝛼0 t1

On substituting, 62.6 / 50 = 1 + 1/234.5 t2 / 1 + 1/ 234.5 x 15

Or 1.25 = 1 + 0.00426 t2 / 1 + 0.00426 x 15 = 1 + 0.00426 t2 / 1.0639

1 + 0.00426 t2 = 1.0639 x 1.25

0.00426 t2 = 1.3298 – 1 = 0.3298

t2 = 0.3298 / 0.00426

= 77.43℃

Example 17; The resistance of a tungsten wire at 20℃ is 1.5 ohm. What will be its resistance at 0℃, 100℃, and 1000℃, given 𝛼 = 0.0046℃ -1 for tungsten?

Solution;

R20 = 1.5-ohm,   t = 20℃,                       𝛼 = 0.0046℃-1,             R0 =?

(a). Using the relation R1 = R0 (1 + 𝛼t)

We get R0 = Rt / 1 + 𝛼t = R20 / 1 + 20𝛼 = 1.5 / 1 + 20 x 0.0046

R0 = 1.5 / 1+ 0.092 = 1.5 / 1.092 = 1.37 Ω

(b). At temperature t = 100℃, we have

Rt = R0 (1 + 𝛼t)

= 1.37 (1 + 0.0046 x 100)

=1.37 x 1.46 = 2.002 Ω

At Rt = R0 (1 + 𝛼t)

= 1.37 (1 + 0.0046 x 1000)

=1.37 (1+4.6)

=1.37 x 5.6 = 7.67 Ω

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