Q.A battery of emf $10\ \text{V}$ and internal resistance $3\ \Omega$ is connected to a resistor. If the current in the circuit is $0.5\ \text{A}$, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Concept: Internal Resistance — the battery itself opposes current flow, so the terminal voltage drops under load.

Step 1 — Total circuit resistance from Ohm’s law:

$R_{\text{total}} = \frac{\mathcal{E}}{I} = \frac{10}{0.5} = 20\ \Omega$.

Step 2 — The total is the sum of external resistor $R$ and internal resistance $r$:

$R + r = 20\ \Omega \quad\Rightarrow\quad R = 20 - 3 = 17\ \Omega$.

Step 3 — Terminal voltage $V$ is the emf minus the drop across $r$:

$V = \mathcal{E} - I r = 10 - (0.5 \times 3) = 10 - 1.5 = 8.5\ \text{V}$.

When a battery is connected in a circuit, it’s not a perfect voltage source. Inside every real battery, there’s a small internal resistance $r$ that opposes the flow of current. This means the voltage you actually measure across the battery’s terminals — called the terminal voltage $V$ — is less than its emf $\mathcal{E}$ when current flows. The difference is the voltage drop across the internal resistance: $V = \mathcal{E} - Ir$.

Here, the battery has $\mathcal{E} = 10\ \text{V}$ and $r = 3\ \Omega$, and the circuit current is $I = 0.5\ \text{A}$. The external resistor $R$ is what we need to find.

1. Apply Ohm’s law to the entire closed circuit. The total resistance in the circuit is $R + r$. The emf drives current through this total resistance, so:

Plug in the known values:

2. Solve for $R$. Multiply both sides by $R + 3$:

3. Find the terminal voltage $V$. The terminal voltage is the voltage across the external resistor (or equivalently, the emf minus the internal drop):

Alternatively, using $V = \mathcal{E} - I r$:

Both methods agree.

Method: Kirchhoff's Voltage Law (or the EMF–Internal Resistance Equation)

This is a standard closed-loop circuit problem. The battery has an emf ($\mathcal{E}$) and an internal resistance ($r$). When current flows, the terminal voltage drops due to the internal resistance.

Steps

1. Write the loop equation For a battery with internal resistance $r$ connected to an external resistor $R$, the total resistance in the circuit is $R + r$. By Kirchhoff's Voltage Law (or Ohm's law for the whole circuit):

2. Substitute known values Given: $\mathcal{E} = 10\ \text{V}$, $r = 3\ \Omega$, $I = 0.5\ \text{A}$

3. Solve for $R$

4. Find terminal voltage Terminal voltage $V$ is the voltage across the external resistor (or the battery's output terminals when current flows):

Quick Check

• Voltage across $R$: $I R = 0.5 \times 17 = 8.5\ \text{V}$ ✓
• Voltage drop across $r$: $I r = 1.5\ \text{V}$
• Sum: $8.5 + 1.5 = 10\ \text{V}$ (matches emf) ✓

Key takeaway: The internal resistance always reduces the terminal voltage below the emf when current flows.

Here are the most common mistakes students make on this classic Internal Resistance problem, and how to avoid each one.

1. Forgetting the Internal Resistance is in Series

The Mistake:

Students often treat the internal resistance ($r$) as separate from the circuit or ignore it entirely. They apply Ohm’s law directly as:

This gives the wrong resistor value.

Why it’s wrong:

The battery’s emf ($\mathcal{E}$) is the total voltage supplied, but some of it is “lost” across the internal resistance. The total resistance in the circuit is $R + r$, not just $R$.

How to Avoid:

Always write the loop equation first:

Then solve for $R$:

Key rule: Internal resistance is always in series with the external load.

2. Confusing Terminal Voltage with Emf

The Mistake:

Students assume the terminal voltage ($V$) equals the emf ($\mathcal{E}$) even when current flows. They answer:

Why it’s wrong:

Terminal voltage is less than emf when current flows because of the voltage drop across $r$:

How to Avoid:

Remember:

• Emf = open-circuit voltage (no current).
• Terminal voltage = voltage across the battery terminals when current flows.

Calculate it explicitly:

Key rule: $V_{\text{terminal}} = \mathcal{E} - I r$ (for a source delivering current).

3. Using the Wrong Formula for Terminal Voltage

The Mistake:

Some students use $V = IR$ (with $R$ as the external resistor) and get:

This gives the correct numerical answer but for the wrong reason — they miss the concept.

Why it’s risky:

If the problem changes (e.g., asks for terminal voltage when the external resistor is unknown), this shortcut fails. It also hides the physics.

How to Avoid:

Always derive terminal voltage from the emf minus internal drop:

Then verify it matches $V = I R$ as a consistency check.

4. Sign Errors in the Loop Equation

The Mistake:

When writing Kirchhoff’s voltage law, students misplace signs — e.g., writing:

Why it’s wrong:

The internal resistance causes a voltage drop in the same direction as the external resistor.

How to Avoid:

Trace the loop in the direction of current:

• Going from negative to positive inside the battery: +$\mathcal{E}$
• Across any resistor (including $r$): $-IR$ or $-Ir$

Correct equation:

Quick Summary Table

Final Tip: Always draw the circuit — battery with $r$ inside, external $R$ outside — and label the terminal voltage across the battery’s external terminals. This visual stops most errors.