Worked examples: sagging terminals and measured r

Two problems — compute the terminal voltage of a real battery under load, then work backward from measurements to find its internal resistance. Both are KVL around a single loop with the internal resistance represented explicitly.

Example 1 — terminal voltage under a heavy load

A battery is labelled $\mathcal{E} = 12\,\text{V}$ and has an internal resistance $r = 0.5\,\Omega$. It drives a load of $R = 3\,\Omega$. Find the terminal voltage and the power delivered to the load.

1. Total loop resistance. $R_{total} = r + R = 0.5 + 3 = 3.5\,\Omega$.
2. Loop current. $I = \mathcal{E} / R_{total} = 12 / 3.5 \approx 3.43\,\text{A}$.
3. Drop across the internal resistance. $V_r = Ir = 3.43 \cdot 0.5 \approx 1.71\,\text{V}$.
4. Terminal voltage. $V_{term} = \mathcal{E} - V_r = 12 - 1.71 \approx 10.29\,\text{V}$. Cross-check: $V_{term} = IR = 3.43 \cdot 3 \approx 10.29\,\text{V}$. ✓
5. Power delivered to the load. $P_R = I^2 R = 3.43^2 \cdot 3 \approx 35.3\,\text{W}$. Power wasted in $r$: $P_r = I^2 r \approx 5.88\,\text{W}$. Efficiency $\eta = P_R / (P_R + P_r) \approx 85.7\%$.

The battery “feels” like an 11 V battery because the published 12 V is trimmed by 1.7 V inside before reaching the load. This is why high-current devices need batteries with low internal resistance — otherwise you waste voltage heating up the cell.

Example 2 — measuring r from two data points

A battery is tested under two different loads. At a 0.5 A draw the terminal voltage reads 9.85 V; at a 2.0 A draw it reads 9.40 V. Find $\mathcal{E}$ and $r$.

1. Set up the two equations. Using $V = \mathcal{E} - Ir$:
   $$9.85 = \mathcal{E} - 0.5 r$$
   $$9.40 = \mathcal{E} - 2.0 r$$
2. Subtract to eliminate $\mathcal{E}$. $9.85 - 9.40 = (2.0 - 0.5) r \Rightarrow 0.45 = 1.5 r$. So $r = 0.30\,\Omega$.
3. Back-substitute for ℰ. $\mathcal{E} = 9.85 + 0.5 \cdot 0.30 = 10.0\,\text{V}$. Check with the second point: $10.0 - 2.0 \cdot 0.30 = 9.40\,\text{V}$. ✓
4. Interpretation. The nameplate is 10 V, the effective internal resistance is 0.30 Ω. At modest loads (< 1 A) the terminal voltage stays within 0.3 V of nameplate; under a 10 A surge it would crash to 10 − 3 = 7 V. The cell's suitability depends entirely on how much current the target device draws.