Voltage Divider Formula: Resistive Voltage Divider, Applications, Questions

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Jasmine Grover

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Voltage Divider Formula is \(V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}\). Here, R1 = Resistor closest to the input voltage (Vin); R2 = Resistor closest to ground; Vin= Input Voltage; Vout = Output voltage across R which is then divided voltage (1/4 of input voltage). Voltage Divider Circuits can be used to generate many voltage levels from a single supply voltage. The common supply can be a single positive or negative supply, such as +5V, +12V, -5V, or -12V, etc. It can also be a dual supply, such as ±5V, ±12V, etc. Since "Volt," the unit of voltage, reflects the amount of potential difference between two places, voltage dividers are also known as potential dividers. 

Key Terms: Volt, Resistor, Potentiometer, Capacitor, Inductors, Voltage Divider, Voltage, Electric current, Ohm’s law, Kirchhoff’s voltage law


Resistive Voltage Divider Circuit

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R1 and R2 are two resistors that are connected in series in this circuit. Because the two resistors are connected in series, an identical amount of electric current must pass through each resistor element of the circuit because there is nowhere else for it to go. As a result, each resistive element has an I*R voltage drop.

We can use Kirchhoff's Voltage Law (KVL) and Ohm's Law to discover the voltage dropped across each resistor calculated in terms of the common current, I, flowing through them with a supply or source voltage, VS, applied across this series combination. 

Voltage Divider Main Circuit

Voltage Divider Main Circuit

Following Ohm's Law, the current flowing across the series network is simply I = V/R. Because the current is shared by both resistors (IR1 = IR2), the voltage dropped across resistor R2 in the aforementioned series circuit can be calculated as follows:

IR2 = VR2/R2 = V/(R1 + R2)

∴ VR2 = VS (R2 / R1 + R2)

Likewise for resistor R1 as being:

IR1 = VR1/R1 = VS/(R1 + R2)

∴ VR1 = VS (R1 / R1 + R2)

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Voltage Divider Formula

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The voltage divider is a series of resistors or capacitors that may be tapped at any point along their length to yield a portion of the voltage applied between their ends.

It comprises a two-resistor electric circuit and a single input voltage supply. A basic voltage divider is depicted in the diagram below. Two resistors are linked in series in this circuit. The voltage divider's output voltage is a function of the input voltage. This circuit aids in determining how the input voltage is distributed among the circuit's components.

A voltage divider divides a voltage between two resistors in a series. We can draw the circuit in a variety of ways, but it should always be the same circuit.

Thus the Voltage Divider Formula is as follows:

\(V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}\)

  • R1 = Resistor closest to the input voltage (Vin)
  • R2 = Resistor closest to ground
  • Vin= Input Voltage
  • Vout = Output voltage across R which is then divided voltage (1/4 of input voltage)

Applications of Voltage Divider

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Voltage Dividers are widely utilized in both analog and digital circuits. The following are some of the most prevalent uses for voltage divider circuits.

Potentiometers

A potentiometer, or simply a POT, is a three-terminal variable resistor. A Voltage Divider circuit with changeable output voltage can be implemented using a potentiometer.  This can be accomplished by connecting the input voltage to the potentiometer's extreme pins and the output to the wiper terminal.

Resistive Sensors (LDR and Thermistor)

LDR (Light Dependent Resistor) and Thermistor are two often used sensors in DIY projects. These two sensors are resistive. The issue is that a Microcontroller, such as Arduino, can only read voltages at the input. You may retrieve the voltage across the resistive sensors (LDR or Thermistor) by putting them in a voltage divider circuit and programming the microcontroller to scale the value accordingly.


Things to Remember

  • Voltage dividers are employed in the measurement of sensors, voltage, logic level shifting, and signal level control.
  • Voltage dividers are used in amplifiers to alter the signal level, measure voltage, and bias active components. Voltage dividers can be found in a multimeter and a Wheatstone bridge.
  • In power transmission, a capacitive voltage divider is used to compensate for load capacitance and to assess high voltage.
  • The voltage divider is only used in circuits when the voltage is regulated by dropping a specific value. It is primarily employed in systems where energy efficiency is not a priority.

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Sample Questions

Ques. Let’s assume the total resistance of a variable resistor is 12 Ω. The sliding contact is positioned at a point where resistance is divided into 4 Ω and 8Ω. The variable resistor is connected across a 2.5 V battery. Let’s examine the voltage that appears across the voltmeter connected across the 4 Ω section of the variable resistor. (2 Marks)

Ans. According to the voltage divider rule, voltage drops will be,

Vout= 2.5Vx4 Ohms/12Ohms=0.83V

Ques. When the two capacitors C1-8uF & C2-20uF are connected in series in the circuit, the RMS voltage drops can be calculated across every capacitor when they are connected to 80Hz RMS supply & 80 volts. (3 Marks)

Ans. Xc1 = 1/ 2πfc1

1/2×3.14x80x8x10-6 = 1/4019.2×10-6

=248.8 ohms

Xc2 = 1/ 2πfc2

1/2×3.14x80x20x10-6 = 1/10048 x10-6

= 99.52 ohms

XCT = XC1 + XC2

= 248.8 + 99.52 = 348.32

VC1 = Vs (XC1/ XCT)

80 (248.8/348.32) = 57.142

VC2 = Vs (XC2/ XCT)

80 (99.52/348.32) = 22.85

Ques. When the two inductors L1-8 mH & L2- 15 mH are connected in series, the RMS voltage drop across every capacitor can be calculated once they are connected to a 40 volts, 100Hz RMS supply. (3 Marks)

Ans. XL1 = 2πfL1

= 2×3.14x100x8x10-3 = 5.024 ohms

XL2 = 2πfL2

= 2×3.14x100x15x10-3

9.42 ohms

XLT = XL1 + XL2

14.444 ohms

VL1 = Vs (XL1/ XLT)

= 40 (5.024/14.444) = 13.91 volts

VL2 = Vs (XL2/ XLT)

= 40 (9.42/14.444) = 26.08 volts

Ques. Find out the output voltage of the voltage divider circuit whose two registers are 6ω and 8ω respectively and the input voltage is 20V. Where 8 ω is in parallel to the output voltage. (2 Marks)

Ans. Vout=8/6+8×20 

= 8/14×20

= 80/7

Vout=11.43V

Therefore the output voltage will be 11.43 V.

Ques. The value of the input voltage of a voltage divider circuit is 20V. The resistors are 5 ω and 7 ω Where 7 ω is in parallel to the output voltage. Compute the output voltage. (2 Marks)

Ans. Vout=7/5+7×20 

= 7/12×20

= 35/3

Vout = 11.66 V

Therefore the output voltage will be 11.66 V.

Ques. Determine the output voltage of the voltage divider circuit whose Ra and Rb are 6 Ω and 8 Ω respectively and the input voltage is 10v. (2 Marks)

Ans. Ra = 6Ω,

Rb = 8Ω

Vin = 10V

= [8 / (6 + 8)] 10

Vout = 5.71V

Ques. The value of the input voltage of a voltage divider is 20V, and the resistors are 5 Ω and 7 Ω. Determine the output voltage. (4 Marks)

Ans. Ra = 5 Ω

Rb = 7 Ω

Vin = 20V

=[ 7 /( 5 + 7 )]20

Vout = 11.66V

Ques. Calculate the output voltages of these two voltage divider circuits (VA and VB):
calculate the voltage between points A (red lead) and B (black lead)
Now, calculate the voltage between points A (red lead) and B (black lead) (VAB).

Ans. VA = + 65.28 V

VB = + 23.26 V

VAB = + 42.02 V (point A being positive relative to point B)

VAB = + 42.02 V (point A being positive relative to point B)

Ques. Current in a series circuit may be calculated with this formula: I = Etotal/Rtotal. And voltage dropped across any single resistor in a series circuit may be calculated with this formula: ER = I R. Combine these two formulae into one, in such a way that the I variable is eliminated, leaving only ER expressed in terms of Etotal, Rtotal, and R. (1 Mark)

Ans. ER = Etotal (R/Rtotal)    

Ques. For a parallel connected resistor R1, R2 and a voltage of V volts. Current across the first resistor is given by
a) I R1
b) I R2
c) I R1 / R1 + R2
d) I R2 / R1 + R2 (3 Marks)

Ans. Explanation: I1 = V / R1

R = R1. R2 / R1 + R2

= I . R1. R2 / R1 . R1 + R2

I1 = I R2 / R1 + R2

Ques. R1 = 1Ω, R2 = 3Ω, R3 = 5Ω and R4 = 7Ω connected in parallel. Total Current = 23A. Then V, I1 , I2 =?
a) 12.26v, 1.725, 2.875
b) 12.23v, 2.875, 1.725
c) 11.26v, 1.95, 1.74
d) 11.23v, 1.74, 1.95 (3 Marks)

Ans. V = I/R

V = I (R1 + R2) R1 R2 = 12.26v

I1 = IR2/ R1 + R2 = 1.725A

I2 = IR1/ R1 + R2

= 2.875A

Ques. A transformer has 500 turns in the primary and 250 turns in the secondary. What is the turn’s ratio? How much is the secondary voltage with a primary voltage of 220V? (3 Marks)

Ans. Np = 500, Ns = 250

Turns Ratio = Ns: Np = 250: 500 = 1: 2

Ns/NP = Vs/VP , 250/500 = Vs/220 

Vs = 110 V


Previous Year Questions 

  1. If now we have to change the null point at 9th  wire, what should we do?… [DUET 2007]
  2. The electrical permittivity and magnetic permeability of free space are​… [DUET 2003]
  3. When the same resistances are connected in series across the same cell, the power developed is…. [KCET 1998]
  4. The resistance between any two terminals is when connected in a triangle is…. [NEET 1993]
  5. potential drop through 4Ω  resistor is… [NEET 1993]
  6. A superconductor exhibits perfect….​ [KCET 2002]
  7. Value of R for which the power delivered in it is maximum is given by... [NEET 1992]
  8. An AC supply gives 30V rms which is fed on a pure resistance of… [JIPMER 2003]
  9. An inductive circuit contains a resistance of… [VITEEE 2011]
  10. An alternating voltage of 220 V, 50 Hz frequency… [VITEEE 2017]
  11. RMS value of AC is _______ of the peak value… [VITEEE 2006]
  12. The instantaneous values of alternating current and voltages… [NEET 2012]
  13. The average power dissipated in… [KCET 2014]
  14. The instantaneous voltage through a device… [KEAM]
  15. The primary of a transformer has 400 turns while the secondary has… 
  16. The average power dissipated in a pure capacitance… [JKCET 2009]
  17. The instantaneous voltages at three terminals marked… [JEE Advanced 2017]
  18. The instantaneous voltage of a… [COMEDK UGET 2015]
  19. 120AC voltage is applied to 1010 ohm resistance. The peak voltage across… 
  20. Phase difference between voltage and current in a capacitor… 

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CBSE CLASS XII Related Questions

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      • 2.
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                          CBSE CLASS XII Previous Year Papers

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