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SS1: PHYSICS – 3RD TERM

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  1. Production of Electric Current | Week 1
    6 Topics
    |
    1 Quiz
  2. Electric Current | Week 2
    5 Topics
    |
    1 Quiz
  3. Electrical Resistance of a Conductor | Week 3
    5 Topics
    |
    1 Quiz
  4. Particulate Nature of Matter | Week 4
    5 Topics
    |
    1 Quiz
  5. Crystalline and Non-crystalline Substances | Week 5
    3 Topics
    |
    1 Quiz
  6. Elastic Properties of Solids | Week 6 & 7
    4 Topics
    |
    1 Quiz
  7. Fluids at Rest & in Motion | Week 8 & 9
    6 Topics
    |
    1 Quiz
  8. Solar Collector
    3 Topics
    |
    1 Quiz



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The electrical resistance of a conductor is affected by the following factors:

  • The temperature of the conductor.
  • The length of the conductor.
  • The cross-sectional area of the conductor.
  • The nature of the material of which the conductor is made.

1. Temperature: The resistance of a conductor increases as the temperature of the conductor increases, i.e.

⇒ \( \scriptsize R \propto T \)

2. Length: The resistance of a conductor is directly proportional to the length; as the length of a conductor increases, the resistance also increases, i.e.

⇒ \( \scriptsize R \propto L \)

3. Cross-sectional Area: The resistance of a conductor increases as the area of the conductor decreases, which reveals that the resistance of a wire or a conductor is inversely proportional to the area, i.e.

⇒ \( \scriptsize R \propto \normalsize \frac{1}{A} \)

4. The Nature: The nature of material of the conductor also affects the resistance of a conductor.

Combining all factors shows that resistance is directly proportional to the length of a conductor and inversely proportional to the area.

⇒ \( \scriptsize R \propto \normalsize \frac{L}{A} \)

∴ \( \scriptsize R = \normalsize \frac{\rho \:\times\: L}{A} \)

\( \scriptsize \rho \) = a constant called resistivity of a material

Resistivity is the resistance, per unit length, of a conductor of unit cross sectional area.

∴ \( \scriptsize \rho = \normalsize \frac{R\:\times\:A}{L} \)

Where R is in ohms, A in m2 and l in m

The unit for \( \scriptsize \rho\) is ohm-metre (Ω-m)

Example 1:

Calculate the resistivity of a wire of length 5m of cross-sectional area 1.4×10-6m2, its resistance is 3.0Ω

Solution:

\( \scriptsize R = \normalsize \frac{\rho L}{A} \)

\( \scriptsize \rho = \normalsize \frac{RA}{L} \)

 = \( \frac{3.0 \; \times \; 1.4 \times 10^{-6}}{5} \)

 = \( \frac{0.0000042}{5} \)

= 0.00000084

= 8.4 X 10-7Ωm

Example 2:

Find the resistance of a wire of lenth 0.85m, radius 0.25mm and resistivity 4.5 x 10-6Ωm

Solution:

Data given in the question:

l = 0.85m

\( \scriptsize \rho = 4.5 \: \times \: 10^{-6} \: \Omega m \)

radius = 0.25mm

The cross-sectional area of a wire is directly proportional to the square of its radius

A = \( \scriptsize \pi r^2 \)

\( \scriptsize R = \normalsize \frac{\rho \:\times\: L}{A} \)

\( \scriptsize R = \normalsize \frac{4.4 \:\times \: 10^{-6} \:\times\: 0.85}{\pi (2.5 \: \times \: 10^{-4})^2} \)

\( \scriptsize R = \normalsize \frac{3.74 \: \times \: 10^{-6}}{1.963 \: \times \: 10^{-7}} \)

= 19.081 ohms

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