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 Previous Year Papers

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