The Internal Resistance Formula, expressed as \( e = I(r + R) \), relates the electromotive force (EMF) of a cell to its internal resistance and the external load resistance. This formula is a core concept in Class 12 Physics (NCERT Chapter 3 — Current Electricity) and appears regularly in CBSE board exams, JEE Main, and NEET. Understanding it helps students calculate terminal voltage, power loss, and cell efficiency. This article covers the formula derivation, a complete formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications.
Key Internal Resistance Formulas at a Glance
Quick reference for the most important internal resistance and EMF formulas.
- EMF equation: \( e = I(r + R) \)
- Terminal voltage: \( V = e – Ir \)
- Internal resistance: \( r = \frac{e – V}{I} \)
- Current in circuit: \( I = \frac{e}{R + r} \)
- Power dissipated internally: \( P_r = I^2 r \)
- Cells in series (EMF): \( e_{\text{eq}} = e_1 + e_2 + \cdots + e_n \)
- Cells in parallel (internal resistance): \( \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2} + \cdots \)
What is the Internal Resistance Formula?
The Internal Resistance Formula describes the opposition offered by a cell or battery to the flow of electric current within itself. Every real cell has two components: an electromotive force (EMF) that drives current, and an internal resistance that opposes it. This internal resistance arises due to the electrochemical nature of the cell’s materials.
According to NCERT Class 12 Physics, Chapter 3 (Current Electricity), when a current \( I \) flows through a circuit containing a cell of EMF \( e \) and internal resistance \( r \) connected to an external resistance \( R \), the governing equation is:
\( e = I(r + R) \)
This formula tells us that the total EMF of the cell drives current through both the internal resistance and the external resistance. The voltage available at the terminals of the cell (terminal voltage) is always less than the EMF when current flows. Internal resistance causes energy loss in the form of heat inside the cell itself. This concept is fundamental in Class 12 CBSE board exams and in competitive exams like JEE Main and NEET, where problems on terminal voltage, cell combinations, and power loss are very common.
Internal Resistance Formula — Expression and Variables
\[ e = I(r + R) \]
This can also be written as:
\[ V = e – Ir \]
where \( V \) is the terminal voltage across the cell’s terminals.
Rearranging for internal resistance directly:
\[ r = \frac{e – V}{I} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( e \) | Electromotive Force (EMF) | Volt (V) |
| \( I \) | Electric Current | Ampere (A) |
| \( r \) | Internal Resistance of Cell | Ohm (Ω) |
| \( R \) | External Load Resistance | Ohm (Ω) |
| \( V \) | Terminal Voltage | Volt (V) |
| \( P_r \) | Power Dissipated Internally | Watt (W) |
Derivation of the Internal Resistance Formula
Consider a cell with EMF \( e \) and internal resistance \( r \) connected to an external resistance \( R \).
Step 1: The total resistance in the circuit is \( R_{\text{total}} = R + r \).
Step 2: By Ohm’s law, the current in the circuit is \( I = \frac{e}{R + r} \).
Step 3: Rearranging gives \( e = I(R + r) = IR + Ir \).
Step 4: The voltage drop across the external resistance is \( V = IR \). Therefore, \( e = V + Ir \), which gives terminal voltage \( V = e – Ir \).
Step 5: Solving for internal resistance: \( r = \frac{e – V}{I} \).
This derivation is directly based on NCERT Class 12 Physics, Chapter 3, and is a standard derivation asked in CBSE board exams.
Complete Electricity Formula Sheet (Current Electricity)
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Internal Resistance Formula | \( e = I(r + R) \) | e=EMF, I=current, r=internal resistance, R=external resistance | V, A, Ω | Class 12, Ch 3 |
| Terminal Voltage | \( V = e – Ir \) | V=terminal voltage, e=EMF, I=current, r=internal resistance | V | Class 12, Ch 3 |
| Current in Circuit | \( I = \frac{e}{R + r} \) | I=current, e=EMF, R=external resistance, r=internal resistance | A | Class 12, Ch 3 |
| Ohm’s Law | \( V = IR \) | V=voltage, I=current, R=resistance | V | Class 12, Ch 3 |
| Power Dissipated Internally | \( P_r = I^2 r \) | P=power, I=current, r=internal resistance | W | Class 12, Ch 3 |
| Power Delivered to External Load | \( P_R = I^2 R \) | P=power, I=current, R=external resistance | W | Class 12, Ch 3 |
| Cells in Series — EMF | \( e_{\text{eq}} = e_1 + e_2 + \cdots + e_n \) | e=EMF of each cell | V | Class 12, Ch 3 |
| Cells in Series — Internal Resistance | \( r_{\text{eq}} = r_1 + r_2 + \cdots + r_n \) | r=internal resistance of each cell | Ω | Class 12, Ch 3 |
| Cells in Parallel — EMF (identical cells) | \( e_{\text{eq}} = e \) | e=EMF of each identical cell | V | Class 12, Ch 3 |
| Cells in Parallel — Internal Resistance | \( r_{\text{eq}} = \frac{r}{n} \) | r=internal resistance of each cell, n=number of cells | Ω | Class 12, Ch 3 |
| Efficiency of Cell | \( \eta = \frac{R}{R + r} \times 100\% \) | R=external resistance, r=internal resistance | % | Class 12, Ch 3 |
| Condition for Maximum Power Transfer | \( R = r \) | R=external resistance, r=internal resistance | Ω | Class 12, Ch 3 |
Internal Resistance Formula — Solved Examples
Example 1 (Class 10 / Class 12 Basic Level)
Problem: A cell has an EMF of 2 V and an internal resistance of 0.5 Ω. It is connected to an external resistance of 3.5 Ω. Find the current flowing through the circuit and the terminal voltage of the cell.
Given: EMF \( e = 2 \) V, internal resistance \( r = 0.5 \) Ω, external resistance \( R = 3.5 \) Ω
Step 1: Write the Internal Resistance Formula for current: \( I = \dfrac{e}{R + r} \)
Step 2: Substitute the values: \( I = \dfrac{2}{3.5 + 0.5} = \dfrac{2}{4} = 0.5 \) A
Step 3: Calculate terminal voltage using \( V = e – Ir \):
\( V = 2 – (0.5 \times 0.5) = 2 – 0.25 = 1.75 \) V
Answer
Current \( I = 0.5 \) A | Terminal Voltage \( V = 1.75 \) V
Example 2 (Class 12 / CBSE Board Level)
Problem: A battery of EMF 12 V delivers a current of 2 A when connected to an external resistance. The terminal voltage measured across the battery terminals is 10.4 V. Find the internal resistance of the battery and the power dissipated inside it.
Given: EMF \( e = 12 \) V, current \( I = 2 \) A, terminal voltage \( V = 10.4 \) V
Step 1: Use the formula for internal resistance: \( r = \dfrac{e – V}{I} \)
Step 2: Substitute values: \( r = \dfrac{12 – 10.4}{2} = \dfrac{1.6}{2} = 0.8 \) Ω
Step 3: Calculate power dissipated internally using \( P_r = I^2 r \):
\( P_r = (2)^2 \times 0.8 = 4 \times 0.8 = 3.2 \) W
Step 4: Verify using \( P_r = (e – V) \times I = 1.6 \times 2 = 3.2 \) W — confirmed.
Answer
Internal Resistance \( r = 0.8 \) Ω | Power dissipated inside battery \( P_r = 3.2 \) W
Example 3 (JEE / NEET Level)
Problem: Three identical cells, each with EMF 1.5 V and internal resistance 0.4 Ω, are connected first in series and then in parallel across an external resistance of 1.2 Ω. Find the current delivered to the external resistance in each case and compare the terminal voltages.
Given: For each cell: \( e = 1.5 \) V, \( r = 0.4 \) Ω; \( R = 1.2 \) Ω; \( n = 3 \) cells
Case A: Cells in Series
Step 1: Equivalent EMF: \( e_{\text{eq}} = 3 \times 1.5 = 4.5 \) V
Step 2: Equivalent internal resistance: \( r_{\text{eq}} = 3 \times 0.4 = 1.2 \) Ω
Step 3: Current: \( I_S = \dfrac{4.5}{1.2 + 1.2} = \dfrac{4.5}{2.4} = 1.875 \) A
Step 4: Terminal voltage (series): \( V_S = I_S \times R = 1.875 \times 1.2 = 2.25 \) V
Case B: Cells in Parallel
Step 5: Equivalent EMF: \( e_{\text{eq}} = 1.5 \) V (same as one cell)
Step 6: Equivalent internal resistance: \( r_{\text{eq}} = \dfrac{0.4}{3} \approx 0.133 \) Ω
Step 7: Current: \( I_P = \dfrac{1.5}{1.2 + 0.133} = \dfrac{1.5}{1.333} \approx 1.125 \) A
Step 8: Terminal voltage (parallel): \( V_P = I_P \times R = 1.125 \times 1.2 = 1.35 \) V
Conclusion: Series combination gives a higher current and higher terminal voltage for this external resistance. When \( R \gg r \), series is preferred; when \( R \ll r \), parallel is preferred.
Answer
Series: \( I_S = 1.875 \) A, \( V_S = 2.25 \) V | Parallel: \( I_P \approx 1.125 \) A, \( V_P = 1.35 \) V
Series combination is more effective here since \( R = r_{\text{series}} \).
CBSE Exam Tips 2025-26
- Always distinguish EMF from terminal voltage. EMF is the open-circuit voltage. Terminal voltage is less than EMF when current flows. Many students confuse the two in numerical problems.
- Learn all three forms of the formula. Practise writing \( e = I(r+R) \), \( V = e – Ir \), and \( r = \frac{e-V}{I} \) from memory. CBSE 2025-26 papers often ask you to derive one form from another.
- Memorise the cell combination rules. For series: EMFs add, internal resistances add. For parallel (identical cells): EMF stays the same, internal resistance divides by \( n \). We recommend writing these as a quick table in your revision notes.
- Practise the 3-mark derivation. The derivation of the Internal Resistance Formula from Kirchhoff’s voltage law is a standard 3-mark question in CBSE Class 12 board exams. Practise it at least five times.
- Check units carefully. EMF and terminal voltage are in Volts. Current is in Amperes. Resistance (internal and external) is in Ohms. A unit mismatch is one of the most common reasons for losing marks.
- Use the efficiency formula for higher-order questions. The formula \( \eta = \frac{R}{R+r} \times 100\% \) is asked in CBSE 2025-26 sample papers. Our experts suggest practising at least two problems on cell efficiency.
Common Mistakes to Avoid
- Mistake 1: Treating EMF as terminal voltage. Students often substitute the EMF value directly as the voltage across the external resistance. The correct approach is to first find the terminal voltage \( V = e – Ir \), then use \( V = IR \) for the external circuit.
- Mistake 2: Ignoring internal resistance in series combinations. When cells are in series, the total internal resistance is \( r_{\text{eq}} = r_1 + r_2 + \cdots \). Forgetting to add internal resistances leads to an overestimated current.
- Mistake 3: Applying the parallel EMF formula to non-identical cells. The simplified formula \( e_{\text{eq}} = e \) for parallel cells only applies when all cells are identical. For non-identical cells, use the weighted formula from Kirchhoff’s laws.
- Mistake 4: Confusing power dissipated internally with total power. Total power from the cell is \( P = eI \). Power dissipated internally is \( P_r = I^2 r \). Power delivered externally is \( P_R = I^2 R \). Always identify which power is asked.
- Mistake 5: Forgetting the condition for maximum power transfer. Maximum power is delivered to the external resistance when \( R = r \). Many students miss this in JEE/NEET application problems.
JEE/NEET Application of Internal Resistance Formula
In our experience, JEE aspirants encounter the Internal Resistance Formula in at least 2–3 questions per paper, either directly or embedded in complex circuit problems. NEET also tests this concept in the context of biological cell membranes and electrochemical cells. Here are the three most common application patterns:
Pattern 1: Finding Current and Terminal Voltage
JEE Main frequently gives a circuit with a cell of known EMF and internal resistance. The question asks for the current, terminal voltage, or power delivered to a specific resistor. The key step is applying \( I = \frac{e}{R+r} \) and then \( V = e – Ir \). These are straightforward single-step problems worth 4 marks each.
Pattern 2: Optimising Cell Combinations
JEE Advanced problems often involve \( m \) rows of \( n \) cells in a mixed (series-parallel) combination. The condition for maximum current through the external resistance is \( R = \frac{nr}{m} \). Students must derive this condition using the Internal Resistance Formula and then optimise. We recommend practising at least three such problems from previous JEE Advanced papers (2018–2024).
Pattern 3: Potentiometer and Internal Resistance Measurement
Both JEE and NEET include the potentiometer experiment for measuring internal resistance. The formula used is \( r = \left(\frac{l_1}{l_2} – 1\right) R \), where \( l_1 \) is the balancing length without the shunt and \( l_2 \) is the balancing length with the shunt resistance \( R \). This is derived directly from the Internal Resistance Formula and is a standard 5-mark question in JEE Advanced. In our experience, students who master this derivation gain a significant scoring advantage.
For further reading on related electric circuit formulas, visit the official NCERT Class 12 Physics Chapter 3 on ncert.nic.in.
FAQs on Internal Resistance Formula
Explore more related formulas on ncertbooks.net: learn about the Electric Flux Formula for understanding Gauss’s law, the Angular Speed Formula for rotational motion, and the Average Acceleration Formula for kinematics. For a complete overview of all Physics formulas, visit our Physics Formulas hub.