Resistance Formula MCQ

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Resistance Formula in an electric circuit is defined as the ratio of the voltage applied to the electric current flowing through the circuit. From Ohm’s Law this can be represented as:

V= I x R

Thus , R = V/ I

Here, V= Voltage Applied in Volts

I= Current in the circuit in Amperes

R = Electrical Resistance, in Ohms

The unit of resistance is given in Ohms and is denoted by the symbol Ω. The resistance formula is also given as 1 Ohm= 1 Volt / 1 Ampere. This article covers important choice-based questions from the concept of resistance covered in the Current Electricity chapter of CBSE Class 12 Physics Syllabus. 

Read More: NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 


Class 12 Resistance Formula MCQs

Q.1 Ohm’s Law hold for which of the following?

  1. Metallic Conductors at High Temperature
  2. Metallic Conductors at Low Temperatures
  3. Diodes
  4. Electrolytes

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A.1. The correct option is B. Metallic Conductors at Low Temperatures

Explanation: At Constant Temperature and Zero magnetic field, the electric current flowing through a metallic conductor is directly proportional to voltage across its ends and inversely proportional to its resistance. 

Q.2. Which of the following is an example of non-Ohmic resistance? 

  1. Tungsten Wire
  2. Diode
  3. Copper Wire
  4. Carbon Resistance

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A.2. The correct option is B. Diode

Explanation: Diode is a non-Ohmic resistance material as it does not follow Ohm’s Law. Diodes provide a near constant voltage even when the electric current is varied and thus has its own characteristics. 

Q.3. Which of the following is Ohms Law?

  1. V= I/R
  2. R= VI
  3. V= IR
  4. None of the above

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A.3. The correct option is C. V= IR

Explanation: Ohm’s Law states that the voltage across a conductor is directly proportional to the current. The constant of proportionality is called as the resistance. Mathematically, the formula is Voltage (V) = Current (I) x (Resistance (R)

Read More: Internal Resistance

Q.4. Determine the resistance if the conductor length is doubled and area of cross-section is kept the same.

  1. Resistance halves
  2. Resistance increases by four times
  3. Resistance doubles
  4. There will be no change in the resistance

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A.4. The correct option is C. Resistance doubles

Explanation: From the question, the new parameter of the conductor will be

Length L = 2L

Area, = A

Resistivty is given by 

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

Therefore the new resistivity would be

\(R= \frac{\rho 2L}{A}\)

From the above equation its clear that, the resistance of the conductor would double when the length is doubled but the area is kept constant. 

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Q.5. If the current of 1 A passes through a resistor, the voltage across the resistor is 10 V.  Determine the current whent he voltage accross the resistor is 8 V.

  1. 18 A
  2. 80 A
  3. 0.8 A
  4. 8 A

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A.5. The correct option is C. 0.8 A

Explanation: From Ohm’s Law, R= V/I

There for in case 1:

I = V/R

= 10/1

=10 A.

In the second case, 

R= V/I

⇒ 10 = 8/I

⇒I = 0.8 A

Q.6. If an electric current of 1 mA flows through a conducting material, determine the number of electrons passing per second through the conductor?

  1. 1.6 x 10-121
  2. 6.25 x 1015
  3. 1.6 x 1017
  4. 6.25 x 1010

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A.6. The correct option is B. 6.25 x 1015

Explanation: I = Q/t — (1)

Here, 

I = 1 mA

= 1 x 10-3 A

t = 1 sec

Total charge passing per second, Q = ne

We know from (1) that Q= I x t

ne = It

⇒e= It/n

= 10-3 x 1/ 1.6 x 10-19

= 6.25 x 1015

Read More: Resistor

Q.7. A resistor of 2.5 Ω has has voltage drop across it of 10 V. Determine the current flowing through it?

  1. 6 A
  2. 10 A
  3. 2 A
  4. 4 A

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A.7. The correct option is D. 4 A

Explanation: From Ohm’s Law

V= IR

⇒I = V/R

⇒ I= 10 / 2.5

= 4 A

Also Check Out:

Q.8. Calculate the resistance when the voltage drop across a given conductor is 8 V and an electric current of 4 A flows though the circuit.  

  1. 3 Ω
  2. 2 Ω
  3. 0.2 Ω
  4. 4 Ω

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A.8. The correct option is B. 2 Ω

Explanation: 

From Ohm’s Law

V= IR

⇒R= V/I

⇒ I= 8 / 4

= 2 Ω

Q.9. SI unit of resistance is given by ________

  1. Amperes
  2. Ohms
  3. Volts / Ampere
  4. Both B & C

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A.9. The correct option is D. Both B & C

Explanation: From Ohm’s Law, the resistance is given by

R= V/I

= V / A

This is also called as Ohms and is denoted by  Ω

Q.10. Which of the following does not obey Ohm’s Law? 

  1. Carbon Resistors
  2. High voltage Circuits
  3. Vacuum tubes
  4. Circuits with low current densities

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A.10. The correct option is C. Vacuum tubes

Explanation: Vacuum tubes have a non-linear circuit wherein the conductivity keeps changing with temperature, current and voltage. When we plot these parameters, we see, that the graph is not a stratight line thus indicating that they donot follow Ohm’s Law. 

Also Check Out:

CBSE CLASS XII Related Questions

  • 1.

    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 $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

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

    • 2.

      A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

        • 80 Wb
        • 16 Wb
        • \(8 \times 10^{-2}\) Wb
        • \(8 \times 10^{-3}\) Wb

      • 3.

        (I) The real image of an object placed between $f$ and $2f$ from a convex lens can be seen on a screen placed at the image location. If the screen is removed, is the image still there? Explain.

        (II) Plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.


          • 4.

            Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.


              • 5.

                A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.


                  • 6.

                    A circular coil of 100 turns and radius $\left(\frac{10}{\sqrt{\pi}}\right)$ cm carrying current of $5.0\,\text{A}$ is suspended vertically in a uniform horizontal magnetic field of $2.0\,\text{T}$. The field makes an angle $30^\circ$ with the normal to the coil. Calculate:

                    (I) the magnetic dipole moment of the coil, and

                    (II) the magnitude of the counter torque that must be applied to prevent the coil from turning.

                      CBSE CLASS XII Previous Year Papers

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