The Voltage Divider Formula gives the output voltage across any resistor in a series circuit, expressed as \( V_{out} = V_{in} imes \dfrac{R_2}{R_1 + R_2} \). This formula is a fundamental concept in Class 10 and Class 12 Physics (NCERT Electricity and Current Electricity chapters). It is equally important for JEE Main, JEE Advanced, and NEET aspirants. This article covers the formula expression, derivation, a complete formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Voltage Divider Formulas at a Glance
Quick reference for the most important voltage divider formulas.
- Two-resistor divider: \( V_{out} = V_{in} imes \dfrac{R_2}{R_1 + R_2} \)
- General n-resistor divider: \( V_k = V_{in} imes \dfrac{R_k}{R_1 + R_2 + \cdots + R_n} \)
- Current in divider circuit: \( I = \dfrac{V_{in}}{R_1 + R_2} \)
- Voltage across R1: \( V_1 = V_{in} imes \dfrac{R_1}{R_1 + R_2} \)
- KVL check: \( V_1 + V_2 = V_{in} \)
- Power dissipated: \( P = \dfrac{V_{in}^2}{R_1 + R_2} \)
- Loaded divider output: \( V_{out} = V_{in} imes \dfrac{R_2 \| R_L}{R_1 + R_2 \| R_L} \)
What is the Voltage Divider Formula?
The Voltage Divider Formula describes how an input voltage is distributed across resistors connected in series. When two or more resistors are placed in series and a voltage is applied across the combination, each resistor “drops” a portion of the total voltage. The fraction dropped by each resistor is proportional to its resistance value.
This concept is introduced in NCERT Class 10 Science, Chapter 12 (Electricity), and revisited in greater depth in NCERT Class 12 Physics, Chapter 3 (Current Electricity). The voltage divider is one of the simplest yet most widely used circuits in electronics and electrical engineering.
The rule follows directly from Ohm's Law and Kirchhoff's Voltage Law (KVL). Since the same current flows through all resistors in a series circuit, the voltage across each resistor is proportional to its resistance. For a two-resistor divider with resistances \( R_1 \) and \( R_2 \) and an input voltage \( V_{in} \), the output voltage across \( R_2 \) is given by the Voltage Divider Formula: \( V_{out} = V_{in} imes \dfrac{R_2}{R_1 + R_2} \).
Understanding this formula helps students analyse complex resistor networks, design sensor circuits, and solve CBSE board and competitive exam problems efficiently.
Voltage Divider Formula — Expression and Variables
For a basic two-resistor voltage divider, the formula for the output voltage across \( R_2 \) is:
\[ V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2} \]
Similarly, the voltage across \( R_1 \) is:
\[ V_1 = V_{in} \times \frac{R_1}{R_1 + R_2} \]
The current flowing through the series circuit is:
\[ I = \frac{V_{in}}{R_1 + R_2} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( V_{out} \) | Output voltage (across R₂) | Volt (V) |
| \( V_{in} \) | Input (supply) voltage | Volt (V) |
| \( V_1 \) | Voltage across R₁ | Volt (V) |
| \( R_1 \) | First resistor (series) | Ohm (Ω) |
| \( R_2 \) | Second resistor (output resistor) | Ohm (Ω) |
| \( I \) | Current through the circuit | Ampere (A) |
| \( R_L \) | Load resistance (for loaded divider) | Ohm (Ω) |
| \( P \) | Total power dissipated | Watt (W) |
Derivation of the Voltage Divider Formula
Consider two resistors \( R_1 \) and \( R_2 \) connected in series across a supply voltage \( V_{in} \).
Step 1: The total resistance in the series circuit is \( R_{total} = R_1 + R_2 \).
Step 2: By Ohm's Law, the current through the circuit is \( I = \dfrac{V_{in}}{R_1 + R_2} \).
Step 3: Since the same current \( I \) flows through \( R_2 \), the voltage across \( R_2 \) is:
\[ V_{out} = I \times R_2 = \frac{V_{in}}{R_1 + R_2} \times R_2 = V_{in} \times \frac{R_2}{R_1 + R_2} \]
Step 4: Verification using KVL: \( V_1 + V_{out} = V_{in} \times \dfrac{R_1}{R_1+R_2} + V_{in} \times \dfrac{R_2}{R_1+R_2} = V_{in} \). This confirms the formula is correct.
Complete Physics Formula Sheet — Circuits & Electricity
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Voltage Divider (2 resistors) | \( V_{out} = V_{in} \times \dfrac{R_2}{R_1+R_2} \) | V₀₱ₜ = output voltage, R₁, R₂ = resistances | V | Class 12, Ch 3 |
| Ohm's Law | \( V = IR \) | V = voltage, I = current, R = resistance | V, A, Ω | Class 10, Ch 12 |
| Series Resistance | \( R_s = R_1 + R_2 + \cdots + R_n \) | R₁, R₂, …, Rₙ = individual resistances | Ω | Class 10, Ch 12 |
| Parallel Resistance | \( \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \) | R₁, R₂ = parallel resistances | Ω | Class 10, Ch 12 |
| Kirchhoff's Voltage Law | \( \sum V = 0 \) (closed loop) | V = voltage drops around a loop | V | Class 12, Ch 3 |
| Power Dissipation | \( P = I^2 R = \dfrac{V^2}{R} \) | P = power, I = current, R = resistance, V = voltage | W | Class 10, Ch 12 |
| Loaded Voltage Divider | \( V_{out} = V_{in} \times \dfrac{R_2 \| R_L}{R_1 + R_2 \| R_L} \) | R₂ ∥ Rₗ = parallel combination of R₂ and load Rₗ | V | Class 12, Ch 3 |
| Current Divider Rule | \( I_1 = I_{total} \times \dfrac{R_2}{R_1+R_2} \) | I₁ = current through R₁ in parallel circuit | A | Class 12, Ch 3 |
| EMF and Internal Resistance | \( V = \varepsilon – Ir \) | ε = EMF, I = current, r = internal resistance | V | Class 12, Ch 3 |
| Wheatstone Bridge Balance | \( \dfrac{R_1}{R_2} = \dfrac{R_3}{R_4} \) | R₁, R₂, R₃, R₄ = bridge resistances | Ω | Class 12, Ch 3 |
Voltage Divider Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: Two resistors, \( R_1 = 4\,\Omega \) and \( R_2 = 6\,\Omega \), are connected in series across a 20 V battery. Find the voltage across \( R_2 \).
Given: \( R_1 = 4\,\Omega \), \( R_2 = 6\,\Omega \), \( V_{in} = 20\,\text{V} \)
Step 1: Write the Voltage Divider Formula: \( V_{out} = V_{in} \times \dfrac{R_2}{R_1 + R_2} \)
Step 2: Substitute the values: \( V_{out} = 20 \times \dfrac{6}{4 + 6} \)
Step 3: Simplify: \( V_{out} = 20 \times \dfrac{6}{10} = 20 \times 0.6 = 12\,\text{V} \)
Step 4: Verify using KVL: \( V_1 = 20 \times \dfrac{4}{10} = 8\,\text{V} \). Check: \( 8 + 12 = 20\,\text{V} \). Correct!
Answer
The voltage across \( R_2 \) is 12 V.
Example 2 (Class 11-12 Level)
Problem: A voltage divider circuit uses \( R_1 = 10\,\text{k}\Omega \) and \( R_2 = 15\,\text{k}\Omega \) connected to a 9 V supply. A load resistance \( R_L = 30\,\text{k}\Omega \) is connected across \( R_2 \). Find the output voltage across the load.
Given: \( R_1 = 10\,\text{k}\Omega \), \( R_2 = 15\,\text{k}\Omega \), \( R_L = 30\,\text{k}\Omega \), \( V_{in} = 9\,\text{V} \)
Step 1: Find the parallel combination of \( R_2 \) and \( R_L \):
\[ R_2 \| R_L = \frac{R_2 \times R_L}{R_2 + R_L} = \frac{15 \times 30}{15 + 30} = \frac{450}{45} = 10\,\text{k}\Omega \]
Step 2: Apply the loaded Voltage Divider Formula:
\[ V_{out} = V_{in} \times \frac{R_2 \| R_L}{R_1 + R_2 \| R_L} = 9 \times \frac{10}{10 + 10} = 9 \times 0.5 = 4.5\,\text{V} \]
Step 3: Compare with unloaded output: \( V_{out,unloaded} = 9 \times \dfrac{15}{25} = 5.4\,\text{V} \). The load reduces the output voltage from 5.4 V to 4.5 V.
Answer
The output voltage across the load is 4.5 V. The load effect reduces the output by 0.9 V compared to the unloaded circuit.
Example 3 (JEE/NEET Level)
Problem: In a voltage divider circuit, \( R_1 \) and \( R_2 \) are in series across a 12 V supply. The output is taken across \( R_2 \). When a voltmeter of internal resistance \( 20\,\text{k}\Omega \) is connected across \( R_2 \), it reads 4 V. When the voltmeter is removed, the true voltage across \( R_2 \) is 6 V. Find the values of \( R_1 \) and \( R_2 \).
Given: \( V_{in} = 12\,\text{V} \), \( R_{meter} = 20\,\text{k}\Omega \), \( V_{measured} = 4\,\text{V} \), \( V_{true} = 6\,\text{V} \)
Step 1: From the unloaded condition, use the Voltage Divider Formula:
\[ \frac{R_2}{R_1 + R_2} = \frac{6}{12} = 0.5 \implies R_1 = R_2 \quad \cdots (1) \]
Step 2: Under loaded condition, the parallel combination of \( R_2 \) and \( R_{meter} \) gives:
\[ R_{eff} = \frac{R_2 \times 20}{R_2 + 20} \text{ (k}\Omega\text{)} \]
Step 3: Apply the divider formula with load:
\[ \frac{R_{eff}}{R_1 + R_{eff}} = \frac{4}{12} = \frac{1}{3} \implies 3R_{eff} = R_1 + R_{eff} \implies 2R_{eff} = R_1 \]
Step 4: Substitute \( R_1 = R_2 \) from equation (1):
\[ 2 \times \frac{R_2 \times 20}{R_2 + 20} = R_2 \implies 40 = R_2 + 20 \implies R_2 = 20\,\text{k}\Omega \]
Step 5: From equation (1): \( R_1 = R_2 = 20\,\text{k}\Omega \).
Answer
\( R_1 = R_2 = \mathbf{20\,\text{k}\Omega} \). This problem illustrates how a voltmeter's internal resistance loads the divider and reduces the measured voltage.
CBSE Exam Tips 2025-26
- Always verify with KVL: After applying the Voltage Divider Formula, check that \( V_1 + V_2 = V_{in} \). This quick check earns full marks in CBSE step-marking.
- State the formula first: In 3-mark and 5-mark questions, write the Voltage Divider Formula before substituting values. CBSE awards one mark for writing the correct formula.
- Unit consistency: Keep all resistances in the same unit (either all in Ω or all in kΩ) before substituting. Mixed units are a common error that costs marks.
- Loaded vs. unloaded: The 2025-26 CBSE syllabus includes practical circuit questions. We recommend practising both loaded and unloaded divider problems.
- Diagram marks: Draw the circuit diagram neatly with \( R_1 \), \( R_2 \), \( V_{in} \), and \( V_{out} \) labelled. CBSE awards 1 mark for a correct circuit diagram in current electricity questions.
- Link to Wheatstone bridge: Our experts suggest connecting the voltage divider concept to the Wheatstone bridge, as CBSE often asks comparison questions in Section D.
Common Mistakes to Avoid
- Swapping R₁ and R₂: Students often write \( V_{out} = V_{in} \times \dfrac{R_1}{R_1+R_2} \) when the output is across \( R_2 \). Remember: the formula uses the resistance across which the output is measured in the numerator.
- Ignoring load resistance: In real circuits, connecting a device across \( R_2 \) changes the effective resistance. Always check whether a load is present before using the simple two-resistor formula.
- Using parallel formula for series circuit: The voltage divider applies to a series circuit. Do not confuse it with the current divider, which applies to a parallel circuit.
- Incorrect unit conversion: Mixing Ω and kΩ in the same calculation leads to wrong answers. Convert all values to the same unit before substituting.
- Forgetting that current is the same in series: The entire derivation relies on the fact that \( I \) is identical through \( R_1 \) and \( R_2 \). Students sometimes apply different currents to each resistor, which is incorrect for a series circuit.
JEE/NEET Application of the Voltage Divider Formula
In our experience, JEE aspirants encounter the Voltage Divider Formula in at least 2-3 problems per paper, either directly or embedded in complex network analysis. Here are the key application patterns:
Pattern 1: Voltmeter and Ammeter Loading
JEE Main frequently tests how a voltmeter's finite resistance alters the divider output. The voltmeter acts as \( R_L \) in parallel with \( R_2 \). You must apply the loaded divider formula to find the measured voltage and compare it with the true voltage. This tests your understanding of instrument errors.
Pattern 2: Potentiometer as a Voltage Divider
The potentiometer, covered in NCERT Class 12 Chapter 3, is essentially a continuously variable voltage divider. JEE Advanced problems ask you to find the balance length, which directly uses the voltage divider principle: \( \dfrac{V_{out}}{V_{in}} = \dfrac{l}{L} \), where \( l \) is the balance length and \( L \) is the total wire length. NEET also tests this in the context of comparing EMFs of cells.
Pattern 3: Transistor Biasing Circuits
In JEE Advanced and Class 12 semiconductor questions, the base voltage of a transistor is set using a voltage divider formed by two resistors connected to the supply. The Voltage Divider Formula directly gives the base voltage. Our experts suggest practising at least five such biasing problems before the exam.
For NEET, the voltage divider appears in questions on measuring instruments, potentiometers, and Wheatstone bridge variants. The formula is not computationally heavy, but conceptual clarity is essential. Focus on understanding why the output voltage is proportional to \( R_2 \) and not \( R_1 \).
FAQs on Voltage Divider Formula
Explore more related topics on ncertbooks.net to strengthen your understanding of circuit analysis. Read our detailed article on the Heat Transfer Formula for thermodynamics problems. Check out the Photon Energy Formula for modern physics and NEET preparation. For wave-based questions, visit our guide on the Wave Power Formula. For a complete overview, visit the Physics Formulas hub. You can also refer to the official NCERT website for Class 12 Physics textbook resources.