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

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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. 0.2 

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A.8. The correct option is B. 2 

Explanation: 

From Ohm’s Law

V= IR

⇒R= V/I

⇒ I= 8 / 4

= 2

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  

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.
    The energy of an electron in an orbit in hydrogen atom is \( -3.4 \, \text{eV} \). Its angular momentum in the orbit will be:

      • \( \dfrac{3h}{2\pi} \)
      • \( \dfrac{2h}{\pi} \)
      • \( \dfrac{h}{\pi} \)
      • \( \dfrac{h}{2\pi} \)

    • 2.

      The electric potential (V ) and electric field (⃗ E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the

        • Production of AC is economical.
        • AC can be easily and efficiently converted from one voltage to another.
        • AC can be transmitted economically over long distances.
        • AC is less dangerous.

      • 3.
        The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

          • P
          • Q
          • R
          • S

        • 4.
          Four long straight thin wires are held vertically at the corners A, B, C and D of a square of side \( a \), kept on a table and carry equal current \( I \). The wire at A carries current in upward direction whereas the current in the remaining wires flows in downward direction. The net magnetic field at the centre of the square will have the magnitude:

            • \( \dfrac{\mu_0 I}{\pi a} \) and directed along OC
            • \( \dfrac{\mu_0 I}{\pi a \sqrt{2}} \) and directed along OD
            • \( \dfrac{\mu_0 I \sqrt{2}}{\pi a} \) and directed along OB
            • \( \dfrac{2\mu_0 I}{\pi a} \) and directed along OA

          • 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 \( \sqrt{2} \), and \( \sqrt{3} \).


              • 6.
                Consider a cylindrical conductor of length \( l \) and area of cross-section \( A \). Current \( I \) is maintained in the conductor and electrons drift with velocity \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]

                  CBSE CLASS XII Previous Year Papers

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