What is the electrical resistance of an object?

Electrical Resistance can be defined as the resistance towards the current when it moves via a bulb or any conductor. Electrical resistance is represented by the alphabet R. The unit of resistance is Ohm (Ω).

Step 1: Ohm’s Law

According to Ohm's law,

V ∝ I

\(\therefore\) V = IR

Here,

  • V = Potential Difference (in Volts).
  • I = Current (in Amperes)
  • R = Resistance (in Ohms).

Step 2: Factors That Affect Electrical Resistance

Electrical resistance is proportional to a conductor's length and is also inversely proportional to its cross-sectional area.

\(R = \rho {{L} \over A}\)

Where,

  • L = Length.
  • A = Cross-sectional Area
  • ρ = Resistivity in Ohm meter (Ωm).

Some factors that affect electrical resistance include:

  1. The cross-sectional area, length, and material of the given conductor.
  2. The conducting material's temperature.

Hence, an object’s electrical resistance is directly proportional to the length and is also inversely proportional to the cross-sectional area.


Related Questions:

  1. A battery of 9 V is connected in series with resistors of 0.2 Ohms, 0.3 Ohms, 0.4 Ohms, 0.5Ohms and 12Ohms respectively. How much current would flow through the 12 ohm resistors?
  2. How Do You Determine The Resistance Of A Resistor?
  3. Why is the series arrangement not used for domestic circuits?
  4. Why is the curve representing Ohm's law linear?
  5. How is voltage related to resistance?
  6. An electric bulb is connected to a 220V power supply line, if the bulb draws a current of 0.5A, calculate the power of the bulb.
  7. The resistance R = V/I, where V=100 ± 5.0V and I = 10 ± 0.2A. What is the total error in R?
  8. What is the necessary condition for a conductor to obey Ohm's Law?
  9. State Ohms law. How can it be verified experimentally?
  10. What are the 3 forms of Ohm's law
  11. What Is Ohm's Law Graph?

Read More:

CBSE CLASS XII Related Questions

  • 1.
    Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.


      • 2.
        A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

          • 5 V
          • 3.5 V
          • 2.5 V
          • Zero V

        • 3.
          If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.


            • 4.
              Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

                • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
                • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
                • Assertion (A) is true, but Reason (R) is false.
                • Both Assertion (A) and Reason (R) are false.

              • 5.
                Write any two features of nuclear forces.


                  • 6.
                    Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is \( 4 \, \mu\text{F} \). Calculate the potential difference across the plates of X and Y.

                      CBSE CLASS XII Previous Year Papers

                      Comments


                      No Comments To Show