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.
    Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

      • attract with a force \( \frac{F}{2} \)
      • repel with a force \( \frac{F}{2} \)
      • repel with a force \( F \)
      • attract with a force \( F \)

    • 2.
      A long solenoid of length \( L \) and radius \( r_1 \) having \( N_1 \) turns is surrounded symmetrically by a coil of radius \( r_2 \, (r_2>r_1) \) having \( N_2 \) turns (\( N_2 \ll N_1 \)) around its mid-point. Derive an expression for the mutual inductance of solenoid and coil. Is \( M_{12} = M_{21} \) valid in this case?


        • 3.
          A tank is filled with a liquid to a height of \( 12.5 \, \text{m} \). The apparent depth of a needle lying at the bottom of the tank is measured to be \( 9.0 \, \text{m} \). Calculate the speed of light in the liquid.


            • 4.
              Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]


                • 5.
                  Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))


                    • 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

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