Potential Divider in DC Circuits: Formula, Examples & Applications (Physics Guide)

20.7 POTENTIAL DIVIDER

20.7.1 Potential Divider Formed by Resistors

A potential divider may consist of the following:

(a) Several resistors connected in series

(i) Two resistors are connected in series. This series combination is then connected to a battery of emf E and internal resistance r, as shown in Figure 20.43.

Figure 20.43 Two resistors connected in series are attached to a battery with emf E and internal resistance, r, forming a complete circuit.

(ii) Suppose a p.d. V is set up across the series combination (in this case V is the terminal p.d.). Then the p.d. V₁ across resistance R₁ will be a certain fraction of V and the p.d. V₂ across R₂ another fraction of V such that

$V = V_1 + V_2$

(iii) To determine the fraction of V across R₁, we use

$V_1 = IR_1$

where I is the current flowing through the resistances. But we have

$E = I(r + R_1 + R_2)$

Hence

$\frac{V_1}{E}= \frac{R_1}{r + R_1 + R_2}$

or,

$\boxed{V_1= \left( \frac{R_1}{r + R_1 + R_2} \right)E}$

If r << R₁ and R₂, the internal resistance may be neglected. We have

$\boxed{V_1\cong \left( \frac{R_1}{R_1 + R_2} \right)E}$

Similarly, the fraction of V across R₂ can be shown to be given by the equation

$\boxed{V_2= \left( \frac{R_2}{r + R_1 + R_2} \right)E}$

Note: In general, other than just resistors, the components in series may include capacitors and inductors.

(b) A rheostat connected to a battery

A rheostat can be connected in the manner as shown in Figure 20.44 to act as a potential divider. A fraction of the p.d. V existing across the rheostat can be obtained between one end of the rheostat and the sliding arm. The p.d. V₁ is a fraction of V.

Figure 20.44 A rheostat connected as a potential divider, where a fraction of the total potential difference V is obtained between one end and the sliding arm, producing an output voltage $V_1$

20.7.2 Resistive Load Applied to Potential Divider Circuit

If a resistive load of resistance R is connected parallel to R₁ (or R₂) in Figure 20.43, then the equation for V₁ given above has to be modified. R₁ has to be replaced with R_T, where

$\frac{1 }{R_T}= \frac{1}{R_1}+\frac{1}{R_2}$

However, if the resistance of the load is very large compared with R₁, i.e., R >> R₁, then

$R_T \cong R_1$

Then,

$\boxed{V_1\cong \left( \frac{R_1}{R_1 + R_2} \right)E}$

Example Problems: Potential Divider in DC Circuits (Step-by-Step Solutions)

EXAMPLE 20.9

Refer to the circuit shown in Figure 20.43. It is given that E = 9.0 V, r = 1.0 Ω, R₁ = 5.0 kΩ and R₂ = 13 kΩ. Determine the p.d. across R₁.

Solution

Given:

• Electromotive force, $E = 9.0 \ V$
• Internal resistance, $r = 1.0 \ \Omega$
• Resistance $R_1 = 5.0 \ k \Omega$
• Resistance $R_2 = 13 \ k \Omega$

Concept:

In a potential divider circuit, the voltage across a resistor is a fraction of the total voltage. When internal resistance is present, the voltage across R₁ is given by:

$V_1= \left( \frac{R_1}{r + R_1 + R_2} \right)E$

Answer

Substitute the given values into the formula:

$V_1= \left( \frac{5}{5+13} \right)(9.0)$

$V_1= \left( \frac{5}{18} \right)(9.0)$

$V_1= 2.5 \ V$

The potential difference across R₁ is 2.5 V.

EXAMPLE 20.10

Figure 20.45 A circuit with a fixed resistor across AB and a rheostat BC connected to a battery of emf 12.0 V. The rheostat acts as a potential divider to vary the voltage across AB.

Refer to Figure 20.45. The battery has emf 12.0 V and negligible internal resistance. The resistor across AB has resistance of 50.0 Ω. The rheostat BC has resistance of 10.0 Ω. Determine the range of voltages which is available across AB.

Answer

Given:

• Electromotive force, $E = 12.0 \ V$
• Internal resistance is negligible
• Resistance across AB, $R_{AB} = 50.0 \ \Omega$
• Resistance of rheostat BC, $R_{BC} = 10.0 \ \Omega$

Concept:

This circuit acts as a potential divider using a rheostat. The voltage across AB depends on the position of the sliding contact of the rheostat. By changing the position, the effective resistance in series with R_AB changes, which alters the voltage distribution.

The potential difference across a resistor in a potential divider is given by:

$V = \left( \frac{R}{{Rtotal}} \right)E$

Two extreme positions of the sliding arm must be considered to determine the voltage range.

Solution:

Case 1: Sliding arm at B

When the sliding arm is at B, there is no additional resistance from the rheostat in series. Thus, the entire voltage appears across AB:

$V_1 = \left( \frac{R_{AB}}{{R_{AB}}} \right)E$

$V_1 = \left( \frac{50}{50} \right)(12.0)=12.0 \ V$

Case 2: Sliding arm at C

When the sliding arm is at C, the full resistance of the rheostat (R_BC) is in series with R_AB. The circuit becomes a potential divider:

$V_2 = \left( \frac{R_{AB}}{{R_{AB}}+R_{BC}} \right)E$

$V_2 = \left( \frac{50}{50+10} \right)(12.0)$

$ = \left( \frac{50}{60} \right)(12.0)=10.0 \ V$ 

Conclusion:

The voltage across AB can vary depending on the position of the sliding arm.

Range of voltage across AB: 10 V to 12.0 V

EXAMPLE 20.11

Figure 20.46 A potential divider circuit with a load resistor

The resistors shown in Figure 20.46 have resistances R₁ = 20.0 Ω and R₂ = 40.0 Ω. The battery has emf 12.0 V and negligible internal resistance. A resistive load X of 30.0 Ω is connected across AB. Determine the p.d. across AB.

Answer

Given:

• Resistance $R_1 = 20 \ \Omega$
• Resistance $R_2 = 40 \ \Omega$
• Load resistance $R_{load} = 30 \ \Omega$
• Electromotive force, $E = 12.0 \ V$
• Internal resistance is negligible

Concept:

The load resistance R_load is connected in parallel with R₁. When two resistors are connected in parallel, their effective resistance R_x is given by:

$\frac{1}{R_x} = \frac{1}{R_1} + \frac{1}{R_{load}}$

After finding the equivalent resistance R_x, the circuit becomes a simple potential divider between R_x and R₂. The voltage across AB (V_x) is then given by:

$V_x=\left(\frac{R_x}{R_x+R_2}\right)E$

Step 1: Calculate the effective resistance R_x

$\frac{1}{R_x}= \frac{1}{20} + \frac{1}{30}$

$=\frac{3+2}{60}=\frac{5}{60}$

$R_x=\frac{60}{5}=12.0 \ \Omega$

Step 2: Calculate the potential difference V_x across AB

$V_x=\left(\frac{R_x}{R_x+R_2}\right)E$

$V_x=\left(\frac{12}{12+40}\right)(12.0)$

$=\left(\frac{12}{52}\right)(12.0)$

$= 2.8 \ V$

Answer:

The potential difference across AB is 2.8 V.

EXAMPLE 20.12

Refer to Example 20.11. A resistive load Y of resistance 10.0 Ω is connected across R₂, as shown in Figure 20.47. Determine the p.d. across AB.

Figure 20.47 A potential divider circuit where a load resistor Y is connected in parallel with $R_2$ to determine the potential difference across AB.

Answer

Given:

• Resistance $R_1 = 20 \ \Omega$
• Resistance $R_1 = 40 \ \Omega$
• Load resistance Y, $R_{load} = 10 \ \Omega$
• Electromotive force, $E = 12.0 \ V$
• From Example 20.11: $R_x = 12.0 \ \Omega$
• Internal resistance is negligible

Concept:

The load resistance R_load is connected in parallel with R₂. When two resistors are connected in parallel, the effective resistance R_y is given by:

$\frac{1}{R_y}= \frac{1}{R_2} + \frac{1}{R_{load}}$

After finding R_y, the circuit becomes a potential divider consisting of R_x and R_y. The potential difference across AB is then determined using the voltage divider formula:

$V_{AB}=\left(\frac{R_x}{R_x+R_y}\right)E$

Step 1: Calculate the effective resistance R_y

$\frac{1}{R_y}= \frac{1}{40} + \frac{1}{10}$

$\frac{1}{R_y}= \frac{1+4}{40}= \frac{5}{40}$

$R_y = 40/5 = 8.0 \ \Omega$

Step 2: Calculate the potential difference V_AB

$V_{AB}=\left(\frac{R_x}{R_x+R_y}\right)E$

$V_{AB}=\left(\frac{12}{12+8}\right)(12.0)$

$=\left(\frac{12}{20}\right)(12.0)$

$= 7.2 \ V$

The potential difference across AB is 7.2 V.

EXAMPLE 20.13

Figure 20.48 A complex resistor network connected to a 12.0 V battery with negligible internal resistance, used to determine the potential difference across AB.

Refer to Figure 20.48. The battery has emf 12.0 V and negligible internal resistance. All the values of the external resistances are in ohms. Determine the p.d. across AB.

Answer

Given:

• Electromotive force, E = 12.0 V
• Internal resistance is negligible
• All resistances are in ohms
• Two resistors of 10 Ω are connected in parallel between A and B
• Other resistors: 5 Ω, 15 Ω, 25 Ω, and 22 Ω as shown in the circuit

Concept:

To determine the potential difference across AB, the circuit must be simplified step by step by combining resistors using series and parallel rules:

• For parallel resistors: $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$
• For series resistors: $R = R_1 + R_2$

After simplification, the circuit becomes a potential divider. The voltage across a section is then found using the voltage divider formula:

$V = \left(\frac{R}{R_{total}}\right)E$

Step 1: Combine resistors between A and B

10 Ω || 10 Ω:

$\frac{1}{R_{AB}} = \frac{1}{10} + \frac{1}{10}=\frac{1}{5}$

$R_{AB} = 5 \ \Omega$

Step 2: Combine R_AB with 5 Ω in series

$R_{AC} = 5 + 5 = 10 \ \Omega$

Step 3: Combine R_AC with (15 Ω + 25 Ω)

$\frac{1}{R_{DF}}=\frac{1}{R_{AC}}+\frac{1}{15+25}$

$=\frac{1}{10}+\frac{1}{40}=\frac{1}{8}$

$R_{DF} = 8 \ \Omega$

Step 4: Find p.d. across DF

$V_{DF}=\left(\frac{R_{DF}}{R_{DF}+R_{FG}}\right)E$

$=\left(\frac{8}{8+22}\right)(12)$

$=\left(\frac{8}{30}\right)(12)=3.2 \ V$

Step 5: Find p.d. across AB

$V_{AB}=\left(\frac{R_{AB}}{R_{AB}+R_{BC}}\right)V_{AC}$

$=\left(\frac{5}{5+5}\right)(3.2)$

$=\left(\frac{5}{10}\right)(3.2)=1.6 \ V$

Note: $V_{AC} = V_{DF}$

Answer:

The potential difference across AB is 1.6 V.

Applications of Potential Divider

• Voltage Regulation: A potential divider is used to obtain a desired output voltage from a higher input voltage in electronic circuits.
• Sensor Circuits: It is commonly used with sensors such as thermistors and light-dependent resistors (LDRs) to convert changes in resistance into measurable voltage changes.
• Signal Conditioning: Potential dividers help adjust voltage levels so that signals can be safely processed by electronic devices like microcontrollers.
• Volume Control: In audio devices, a variable resistor (rheostat or potentiometer) acts as a potential divider to control sound levels.
• Biasing of Electronic Components: Used in transistor circuits to set the correct operating voltage for proper functioning.

Conclusion

A potential divider is a simple and effective method for obtaining a desired voltage from a DC source using resistors. By applying basic circuit laws, the output voltage can be accurately determined as a fraction of the input voltage. This concept is widely used in electronic circuits for voltage control, signal processing, and device operation.

Frequently Asked Questions (FAQ)

1. What is a potential divider?

A potential divider is a circuit made of resistors in series used to produce a desired output voltage that is a fraction of the input voltage.

2. How does a potential divider work?

It works by dividing the total voltage across series resistors in proportion to their resistances.

3. What is the formula for a potential divider?

The output voltage is given by V_out = (R₁ / (R₁ + R₂)) V.

4. Where is a potential divider used?

It is used in sensor circuits, voltage regulation, signal conditioning, and electronic device biasing.

5. What happens if a load is connected to a potential divider?

The output voltage changes because the effective resistance is altered, so the formula must be adjusted.

6. Can a variable resistor be used in a potential divider?

Yes, a variable resistor such as a rheostat or potentiometer can be used to adjust the output voltage.

Share :