MCQs On Voltage Divider Formula

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A voltage divider is a resistor or series of resistors or capacitors which is provided with taps at a certain point and used to provide various potential differences from a single power source. A voltage dividers are very useful in providing different voltage levels from a common supply voltage and the common supply is a single supply either positive or negative.


MCQs On Voltage Divider Formula 

Ques. Find the value displayed by a voltmeter in the given image. (1 Mark)

  1. 1V
  2. 1A
  3. 4A
  4. 4V

​ Find the value displayed by a voltmeter in the given image

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Ans. a. 1V

Explanation: Voltage across is 1KΩ is;

V1 = R1/ R1+R2 *V

Putting value in the formula:

V2 = 1/1+4 * 5 = 1V

Ques. Calculate potential difference across 60-watt lamps; Two 220-volt lamps, the first 60 watt and the second 75 watt are connected to a series of 440-volt supplies. (1 Mark)

  1. 300 volt
  2. 245 volt
  3. 200 volt
  4. 345 volt

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Ans. b 245 volt

Explanation: Given;

P1 - 60W; P2 - 75W

V1 - 200V; V2 - 440V

Ques. Where the current division problem arises (1 Mark)

  1. Series connected resistors
  2. Parallel connected resistors
  3. When resistors are equal
  4. Both series and parallel resistors.

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Ans. d. Both series and parallel resistors.

Explanation: The current division problem arises in both series and parallel resistors.

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Ques. 490Ω is connected across a voltage source Vs = 150V. The internal resistance of a source is 10Ω. Calculate the output voltage across the load. (1 Mark)

  1. 180V
  2. 150V
  3. 147V
  4. 120V

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Ans. C. 147V.

Explanation: V = Vs * RL/RL+RS

V = 150 * 490/ 490+10

V = 150 * 490/ 500

V = 147V

Ques. Evaluate the value of resistor R1 in the given figure. (1 Mark)

  1. 1.3Ω
  2. 1.5Ω
  3. 1.7Ω
  4. 1.9Ω

​Evaluate the value of resistor R1 in the given figure

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Ans. b. 1.5Ω

Explanation: The voltage across R1= 6V

Current through R1 = I=4A

Therefore, 

6V = 4R1

Hence R1 = 6/4 = 1.5Ω

Ques. R1 = 1Ω, R2 = 3Ω, R3 = 5Ω and R4 = 7Ω connected in series. Total voltage = 20V, Current I, V2 =? (1 Mark)

  1. I = 1.16, V2 = 3.72
  2. I = 1.25, V2 = 3.75
  3. I = 1.15, V2 =3.73
  4. I = 1.23, V2 = 3.75

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Ans. d. I = 1.23, V2 = 3.75

Explanation: V = I/R

V = I (R1+R2)

R1 R2 = 12.26V

I1 = 1R2/ R1 = R2 = 1.725A

I2 = IR1/ R1+R2

= 2.875A

Ques. Voltage division is necessary for parallel resistance networks (1 Mark)

  1. False
  2. True

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Ans. a. False

Explanation: In parallel, the connection voltage is the same so division is not required.

Ques. Why is the current division necessary? (1 Mark)

  1. Due to different voltage
  2. Kirchhoff's Law
  3. In parallel current differs
  4. In series current, it is the same

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Ans. c. In parallel current differs

Explanation: because in parallel the current differs from all

Ques. Find i =? (1 Mark)

Find i =?

  1. -1A
  2. -2A
  3. +1A
  4. +2A

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Ans. d. +2A

Explanation: i = 1/1+3(8)

= 2A 

Ques. What is an internal voltage drop in an ideal constant voltage source? (1 Mark)

  1. Zero
  2. High
  3. Low
  4. Moderate

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Ans. a. Zero

Explanation: An ideal voltage source has zero source resistance.

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CBSE CLASS XII Related Questions

  • 1.
    Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.


      • 2.
        A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.


          • 3.
            A charged particle $+q$ in an electric field $\vec{E}$ experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field $\vec{B}$. But this magnetic force is perpendicular to both velocity $\vec{v}$ of the charged particle and the magnetic field $\vec{B}$, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses $m$ and $\frac{m}{2}$ having charges $-q$ and $+2q$ respectively. They are accelerated from rest through the same potential difference $V$ and acquire kinetic energy $K_1$ and $K_2$. Then they enter in a region of uniform magnetic field $\vec{B}$ perpendicular to their velocities.


              • 4.
                An electric field $\vec{E}$ is established across the ends of a cylindrical conductor of length L and area of cross-section A. Discuss how electrons attain an average velocity, independent of time. Hence, obtain a relation between current in the conductor and this ‘average velocity’ of electrons.


                  • 5.
                    Consider the nuclear reaction \( X \to Y + Z \). Let \( M_x \), \( M_y \), and \( M_z \) be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

                      • \( (M_x - M_z)<M_y \)
                      • \( (M_x - M_y)<M_z \)
                      • \( M_x>(M_y + M_z) \)
                      • \( M_x<(M_y + M_z) \)

                    • 6.
                      Write two advantages of reflecting telescope over refracting telescope.

                        CBSE CLASS XII Previous Year Papers

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