Factors on which the Resistance of a Conductor Depends
Note the ammeter reading in an electric circuit with:
- A nichrome wire of length l. Assume the reading = 1 A.
- Nichrome wire of twice the length (2l). Here, ammeter reading decreases to one-half (0.5 A).
- A thicker (larger cross-sectional area) nichrome wire of the same length l. Here, reading is increased. If the area is doubled, reading is also doubled (2 A).
- A copper wire of same length and cross-sectional area as that of first nichrome wire. Here, reading is changed.
- Thus, resistance of a conductor depends on (i) its length (ii) its area of cross-section and (iii) nature of material.
- Resistance of a uniform metallic conductor is directly proportional to its length (R ∝ l) and inversely proportional to the area of cross-section (R ∝ 1/A).
R = ρl/A
- Where ρ (rho) is a constant of proportionality and is called electrical resistivity of the material of conductor. It is a characteristic property of the material.
- SI unit of resistivity is Ω m.
- Resistivity of metals & alloys is very low (Range: 10–8 to 10–6 Ω m). They are good conductors of electricity.
- Resistivity of insulators (rubber, glass etc.) is very high (Range: 1012 to 1017 Ω m).
- Resistance and resistivity vary with temperature.
- Resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. So, they are used in electrical heating devices, like electric iron, toasters etc.
- Tungsten is used for filaments of electric bulbs. Copper & aluminium are used for electrical transmission lines.
Electrical resistivity of some substances at 20°C
| Material | Substance | Resistivity (Ω m) |
|---|---|---|
| Conductors | Silver | 1.60 × 10–8 |
| Copper | 1.62 × 10–8 | |
| Aluminium | 2.63 × 10–8 | |
| Tungsten | 5.20 × 10–8 | |
| Nickel | 6.84 × 10–8 | |
| Iron | 10.0 × 10–8 | |
| Chromium | 12.9 × 10–8 | |
| Mercury | 94.0 × 10–8 | |
| Manganese | 1.84 × 10–6 | |
| Alloys | Constantan (Cu + Ni) | 49 × 10–6 |
| Manganin (Cu + Mn + Ni) | 44 × 10–6 | |
| Nichrome (Ni + Cr + Mn + Fe) | 100 × 10–6 | |
| Insulators | Glass | 1010 – 1014 |
| Hard rubber | 1013 – 1016 | |
| Ebonite | 1015 – 1017 | |
| Diamond | 1012 – 1013 | |
| Paper (dry) | 1012 |
Problem:
- a) How much current will an electric bulb draw from a 220 V source, if the resistance of bulb filament is 1200 Ω?
- b) How much current will an electric heater draw from a 220 V source, if the resistance of heater coil is 100 Ω?
Solution:
- (a) V = 220 V; R = 1200 Ω.
- Current, I = V/R = 220 V / 1200 Ω = 0.18 A.
- (b) V = 220 V; R = 100 Ω.
- Current, I = 220 V / 100 Ω = 2.2 A.
- Thus the current drawn by an electric bulb and electric heater from the same 220 V source is different.
Problem: The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?
Solution:
- Potential difference V = 60 V, current I = 4 A.
- Resistance, R = V/I = 60 V / 4 A = 15 Ω.
- When the potential difference is increased to 120 V the current is given by:
I = V/R = 120 V / 15 Ω = 8 A.
Problem: Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Predict the material of the wire.
Solution:
- Resistance R of the wire = 26 Ω.
- Diameter d = 0.3 mm = 3 × 10–4 m.
- Length l of the wire = 1 m.
- Resistivity of the metallic wire, ρ = (RA/l)
- This is the resistivity of manganese.
= (R π d2/4 l).
= 1.84 × 10–6 Ω m.
Problem: A wire of given material having length l and area of cross-section A has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length l/2 and area of cross-section 2A?
Solution:
- For first wire: R1 = ρl/A = 4 Ω.
- For second wire: length = l/2, area = 2A.
- R2 = ρ(l/2)/(2A) = (ρl/A)/4 = 4 Ω / 4 = 1 Ω.
