V = E - Ir. Why terminal voltage drops under load.

Internal Resistance of a Cell — EMF vs Terminal Voltage

Question

A battery has an EMF of 12 V and internal resistance 2 Ω. When connected to an external resistance of 4 Ω, find: (a) the current in the circuit (b) the terminal voltage of the battery

This is a direct CBSE 2024 Board Exam question — the kind that fetches full marks if you know the formula, and zero if you confuse EMF with terminal voltage.

Solution — Step by Step

EMF ( $E$ ) = 12 V, internal resistance ( $r$ ) = 2 Ω, external resistance ( $R$ ) = 4 Ω.

The total resistance in the circuit is $R + r$ — the internal resistance is in series with the external load. This is the key physical picture.

The driving force is the EMF. The total opposition is $(R + r)$ .

I = \frac{E}{R + r} = \frac{12}{4 + 2} = \frac{12}{6} = 2 \text{ A}

The terminal voltage is what you actually measure across the battery’s terminals — it’s less than EMF because the internal resistance “eats” some voltage.

V = E - Ir = 12 - (2 \times 2) = 12 - 4 = \mathbf{8 \text{ V}}

The voltage across the external resistor should equal the terminal voltage:

V = IR = 2 \times 4 = 8 \text{ V} \checkmark

Both methods give 8 V — this cross-check takes 5 seconds and guarantees your answer.

Why This Works

Think of the battery as two things in series: an ideal voltage source (the EMF, $E$ ) and a small resistor (the internal resistance, $r$ ). When current flows, that internal resistor causes a voltage drop of $Ir$ .

The terminal voltage $V = E - Ir$ is what remains after subtracting this internal drop. When no current flows (open circuit), $I = 0$ , so $V = E$ — the terminal voltage equals the EMF. This is actually how we measure EMF in practice using a potentiometer.

As the external load decreases (more current drawn), the drop $Ir$ increases and terminal voltage falls further. This is why old batteries in a torch still show 1.5 V on a multimeter but can’t light a bulb — the internal resistance has increased with age, so under load the terminal voltage collapses.

I = \frac{E}{R + r}

V_{\text{terminal}} = E - Ir = IR

\text{Power lost internally} = I^2 r

Alternative Method

Instead of using $V = E - Ir$ , directly apply the potential divider idea.

The terminal voltage is the fraction of total EMF that appears across $R$ (not $r$ ):

V = E \times \frac{R}{R + r} = 12 \times \frac{4}{6} = 8 \text{ V}

This works because EMF distributes across $R$ and $r$ in proportion (series circuit). In JEE, this form is sometimes faster when $r$ is not explicitly asked.

Common Mistake

Students write $I = E/R$ instead of $I = E/(R+r)$ , getting $I = 12/4 = 3$ A.

This ignores the internal resistance entirely. The internal resistance is inside the battery — you can’t see it — but current definitely flows through it. Always write $R + r$ in the denominator. In CBSE marking schemes, this wrong current propagates and kills both parts (a) and (b).

When a problem says “a cell of EMF $E$ and internal resistance $r$ ”, the $r$ is always in the denominator of the current formula. If $r = 0$ is mentioned, it’s an ideal cell — only then do you use $I = E/R$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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