A cell of emf 2V internal resistance 0.5Ω connected to 4.5Ω — find current and terminal voltage

Question

A cell of emf $\varepsilon = 2\text{ V}$ and internal resistance $r = 0.5\,\Omega$ is connected to an external resistance $R = 4.5\,\Omega$ . Find: (a) The current flowing through the circuit (b) The terminal voltage of the cell

Solution — Step by Step

A real battery (cell) is modelled as an ideal EMF source ( $\varepsilon$ ) in series with a small internal resistance ( $r$ ). When current flows, some voltage drops across this internal resistance — that’s why the terminal voltage is always less than the EMF when the cell is discharging.

Total resistance in the circuit = $R + r$ (they are in series).

Using Ohm’s law for the complete circuit:

I = \frac{\varepsilon}{R + r} = \frac{2}{4.5 + 0.5} = \frac{2}{5} = \mathbf{0.4\text{ A}}

The current flowing through the circuit is 0.4 A.

The terminal voltage (voltage across the external resistance, also the actual voltage at the battery terminals) is:

V_{\text{terminal}} = \varepsilon - Ir = 2 - (0.4)(0.5) = 2 - 0.2 = \mathbf{1.8\text{ V}}

Alternatively, using Ohm’s law for the external resistor:

V_{\text{terminal}} = IR = 0.4 \times 4.5 = 1.8\text{ V}

Both methods give the same answer — a good cross-check.

Why This Works

The EMF represents the energy per unit charge provided by the chemical reaction in the cell. But as charge flows through the internal resistance $r$ , some energy is lost (as heat inside the cell). The voltage available at the terminals — the terminal voltage — is the EMF minus this internal voltage drop.

V_{\text{terminal}} = \varepsilon - Ir

This equation tells us: when no current flows (open circuit), $V_{\text{terminal}} = \varepsilon$ — the terminal voltage equals the EMF. When current flows (closed circuit), the terminal voltage drops. The higher the current, the greater the drop. This is why a battery’s voltage reading drops when it’s under heavy load.

Alternative Method

The two formulae are equivalent:

$V = \varepsilon - Ir$ (EMF minus internal drop)

$V = IR$ (Ohm’s law for external)

Setting them equal: $\varepsilon - Ir = IR \Rightarrow \varepsilon = I(R + r)$ — which is just the original current formula. Use whichever is more convenient given what’s asked.

CBSE Class 10 and 12 both test this concept. In Class 10, it’s about Ohm’s law and circuits. In Class 12, the internal resistance model is formally introduced. JEE often asks about maximum power transfer — maximum power is delivered to $R$ when $R = r$ (maximum power theorem). Know this condition for JEE.

Common Mistake

Students often use only $V = IR$ with just the external resistance and get $V = 0.4 \times 4.5 = 1.8$ V, which is correct for terminal voltage. The mistake is when they use $V = \varepsilon / R$ (ignoring $r$ ): $V = 2/4.5 = 0.44$ A — wrong current. Always use total resistance ( $R + r$ ) when calculating current. Use $R$ alone only when calculating the voltage across the external resistor.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next