Battery emf & Internal Resistance - Physics 2

1. Under the maximum-power-transfer condition, which of the following statements about the efficiency of power delivery from the battery to $R$ is correct?
   Answer: Efficiency is exactly 50%, because at maximum power transfer ($R = r$), the current is $\mathcal{E}/(2r)$, the power in $R$ is $\mathcal{E}^2/(4r)$, and the total power supplied by the emf is $\mathcal{E}I = \mathcal{E}^2/(2r)$, so exactly half the source power is lost in the internal resistance. When a real battery powers an external resistor, you're always dealing with a voltage divider between the internal resistance $r$ and the load $R$. The key insight is that maximizing power to $R$ and maximizing efficiency are two different — and competing — goals. The maximum power transfer theorem tells you that power delivered to $R$ is greatest when $R = r$. At that condition, the current is $I = \mathcal{E}/(R + r) = \mathcal{E}/(2r)$. The power delivered to the external resistor is $P_R = I^2 R = (\mathcal{E}^2/4r^2)\cdot r = \mathcal{E}^2/(4r)$. The total power supplied by the emf is $P_{total} = \mathcal{E}I = \mathcal{E}^2/(2r)$. Efficiency is therefore $\eta = P_R/P_{total} = \tfrac{1}{2} = 50\%$. Answer B captures this exactly — half the source power is dissipated inside the battery as heat in $r$, and half reaches the load. Answer A is internally contradictory: it correctly computes 50% but then invents a fictional "25% impedance matching overhead," which has no physical basis. There are only two resistors sharing power here. Answer C contains a correct efficiency figure but builds a false conclusion around it. As $r$ increases with battery age, $R_{optimal}$ does shift to match the new $r$, but the efficiency at maximum power transfer is always 50%, regardless of the specific values of $\mathcal{E}$ or $r$. Answer D is backwards. Setting $R = r$ actually maximizes — not minimizes — the internal voltage drop, since a large fraction of $\mathcal{E}$ appears across $r$. Study tip: Remember the trade-off: to push efficiency above 50%, you need $R \gg r$, but then power delivered to $R$ drops. Maximum power and maximum efficiency are mutually exclusive — a classic trap on circuit problems.
2. Which of the following correctly identifies the physical meaning of the y-intercept and x-intercept of the $V_T$ vs. $I$ graph?
   Answer: The y-intercept equals $\mathcal{E}$ (the open-circuit emf) and the x-intercept equals $\mathcal{E}/r$ (the short-circuit current), because $V_T = \mathcal{E} - Ir$ predicts these values at $I = 0$ and $V_T = 0$, respectively. Whenever you see a question about a battery's terminal voltage graph, your anchor equation should be $V_T = \mathcal{E} - Ir$. This is a linear equation in slope-intercept form, where $V_T$ is on the y-axis and $I$ is on the x-axis. Recognizing this structure immediately tells you everything about the intercepts. The slope is $-r$ (negative internal resistance), the y-intercept is $\mathcal{E}$ (set $I = 0$, so $V_T = \mathcal{E}$), and the x-intercept is found by setting $V_T = 0$: $0 = \mathcal{E} - Ir \Rightarrow I = \mathcal{E}/r$. This short-circuit current represents the maximum current the battery could theoretically deliver if the external resistance were zero. Answer A captures both intercepts correctly and explains the physical reasoning behind each — making it the right choice. Answer B incorrectly claims the x-intercept equals $r$. The x-intercept has units of current (amperes), not ohms — this is a unit mismatch and a conceptual error. Answer C replaces $r$ with $R$ in the short-circuit current formula. But when $V_T = 0$, the external resistance is irrelevant — only internal resistance $r$ limits current. Answer D describes a charging battery (where $V_T = \mathcal{E} + Ir$), which is the opposite scenario. A discharging battery loses terminal voltage as current increases, not gains it. Study tip: Memorize $V_T = \mathcal{E} - Ir$ as your go-to battery equation and always treat it like $y = mx + b$. Any graph question about terminal voltage is really just asking you to read the intercepts of this line.
3. Using both measurements, what are the emf $\mathcal{E}$ and internal resistance $r$ of the battery?
   Answer: $\mathcal{E} = 12.15 \text{ V}$ and $r = 0.25 \text{ } \Omega$, obtained by writing $\mathcal{E} - Ir = 11.4$ for discharge and $\mathcal{E} + Ir = 12.9$ for charging, then adding to get $2\mathcal{E} = 24.3 \text{ V}$ (so $\mathcal{E} = 12.15 \text{ V}$) and subtracting to get $2Ir = 1.5 \text{ V}$ (so $r = 1.5/6 = 0.25 \text{ } \Omega$). When a battery is discharging, it drives current outward, so its terminal voltage is reduced by the internal resistance drop: $V_T = \mathcal{E} - Ir$. When a battery is charging, an external source forces current inward (opposing the battery's own current direction), so the terminal voltage is raised above the emf: $V_T = \mathcal{E} + Ir$. Keeping these signs straight is the entire key to this problem. Setting up the two equations correctly gives you $\mathcal{E} - 3r = 11.4$ (discharge) and $\mathcal{E} + 3r = 12.9$ (charging). Adding them eliminates $r$: $2\mathcal{E} = 24.3$, so $\mathcal{E} = 12.15 \text{ V}$. Subtracting the first from the second eliminates $\mathcal{E}$: $6r = 1.5$, so $r = 0.25\ \Omega$. This is exactly what C does — clean, systematic, and correct. A reaches the right numerical values but then undermines them with a fabricated caveat about needing a "correction factor." No such correction is required under the standard battery model; this caveat is a distractor designed to make you doubt a perfectly valid result. B misapplies the discharge formula to both scenarios, treating the charging measurement as if it also follows $\mathcal{E} - Ir$. This produces two contradictory equations, and the "fix" of averaging and doubling is entirely made up — not a real physics technique. D incorrectly identifies the discharge terminal voltage as the emf itself, confusing $V_T$ with $\mathcal{E}$. The terminal voltage and emf are only equal when current is zero. Study tip: Whenever a battery problem involves both charging and discharging, immediately write two separate equations with opposite signs on the $Ir$ term, then solve the system — this two-equation approach always works cleanly.
4. What is the current through $R$, and which battery is being charged?
   Answer: $I = 0.50 \text{ A}$ and battery 2 is being charged, because the net emf is $\mathcal{E}_1 - \mathcal{E}_2 = 4 \text{ V}$ and the total resistance is $r_1 + r_2 + R = 8 \text{ }\ \Omega$, giving $I = 4/8 = 0.50 \text{ A}$, with current forced through battery 2 against its emf. When two batteries oppose each other in series, you must apply Kirchhoff's Voltage Law (KVL) carefully. The key insight is that opposing polarities mean the emfs subtract, not add — the stronger battery "wins" and determines the direction of current flow. Going around the loop, the net emf is $\mathcal{E}_1 - \mathcal{E}_2 = 12 - 8 = 4 \text{ V}$. All resistances are in series, so the total resistance is $r_1 + r_2 + R = 1 + 2 + 5 = 8 \text{ }\Omega$. Applying Ohm's law gives $I = 4/8 = 0.50 \text{ A}$. Since battery 1 is stronger, it drives current in its own preferred direction — but that means current enters battery 2 at its positive terminal and exits at its negative terminal, which is the signature of charging: current is forced against battery 2's natural discharge direction. That makes D correct. Choice A gets the current formula right ($4\text{ V}/8\text{ }\Omega$) but then mysteriously reports $2.5 \text{ A}$ — a arithmetic error — and incorrectly identifies which battery is charged; the logic about current direction is actually sound but paired with wrong numbers. Choice B correctly calculates $I = 0.50 \text{ A}$ but flips the charging identification: battery 1 is the stronger one driving the current, so it is not being charged — battery 2 is. Choice C makes a fundamental conceptual error by adding the emfs as if opposing batteries reinforce each other; they don't — opposing polarities subtract. A reliable strategy: always assign a current direction, write the KVL equation with a sign convention, and remember that a battery being charged has current entering its positive terminal.
5. From these two measurements alone, what is the internal resistance $r$ of the battery, and what current does the $10 \text{ } \Omega$ resistor draw?
   Answer: $r = 1.25 \text{ } \Omega$ and $I = 0.80 \text{ A}$, found by recognizing that the open-circuit voltmeter reading gives $\mathcal{E} = 9.0 \text{ V}$, the current under load is $I = V_2/R = 8.0/10 = 0.80 \text{ A}$, and the voltage drop across the internal resistance is $\mathcal{E} - V_2 = 1.0 \text{ V}$, so $r = 1.0/0.80 = 1.25 \text{ } \Omega$. Whenever you see a battery circuit problem, your first instinct should be to identify the EMF and terminal voltage separately. The EMF ($\mathcal{E}$) is the battery's true voltage with no current flowing; the terminal voltage drops once current flows because of the internal resistance $r$. Here's the key insight: a voltmeter has extremely high resistance, so connecting it alone draws essentially zero current. That first reading, $V_1 = 9.0 \text{ V}$, is your open-circuit measurement — meaning $\mathcal{E} = 9.0 \text{ V}$. Once you attach the $10\ \Omega$ resistor, real current flows and the terminal voltage drops to $V_2 = 8.0 \text{ V}$. The current through the external resistor is $I = V_2/R = 8.0/10 = 0.80 \text{ A}$. The voltage "lost" inside the battery is $\mathcal{E} - V_2 = 1.0 \text{ V}$, so the internal resistance is $r = 1.0/0.80 = 1.25\ \Omega$. That's exactly what answer A describes — it's correct. Answer B makes a conceptual error by treating $V_1$ as a loaded terminal voltage rather than the EMF, computing current as $9.0/10 = 0.90 \text{ A}$. This misidentifies what the open-circuit reading actually represents. Answer C attempts a nonsensical hybrid calculation — correctly finding $I = 0.80 \text{ A}$ but then confusing $V_2$ with the EMF, which forces $\Delta V = 0$, and the final value of $r$ is fabricated. Answer D raises a technically real concern about voltmeter loading, but the problem explicitly states the voltmeter has very high resistance — a standard idealization that makes this objection irrelevant here. Your study tip: always ask yourself "Is this an open-circuit or loaded measurement?" before writing any equation. That single distinction separates $\mathcal{E}$ from terminal voltage and unlocks every internal resistance problem.
6. What is the internal resistance $r$ of the battery?
   Answer: $r = 0.48 \text{ }\ \Omega$, found by computing the equivalent external resistance $R_{eq} = (4 \times 6)/(4+6) = 2.4 \text{ }\ \Omega$, the total current $I = V_T/R_{eq} = 5.0/2.4 \approx 2.08 \text{ A}$, and then applying $r = (\mathcal{E} - V_T)/I = 1.0/2.08 \approx 0.48 \text{ }\ \Omega$. When a battery with internal resistance powers an external circuit, you can think of it as a voltage divider: the emf splits between the internal resistance $r$ and the external load. The key relationships are $V_T = \mathcal{E} - Ir$ and $I = V_T / R_{eq}$, where $R_{eq}$ is the equivalent external resistance. Since $R_1$ and $R_2$ are in parallel, their equivalent resistance is $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{4 \times 6}{4 + 6} = 2.4\ \Omega$. Because the terminal voltage appears across both parallel branches equally, the total current drawn from the battery is $I = \frac{V_T}{R_{eq}} = \frac{5.0}{2.4} \approx 2.08\ \text{A}$. The voltage drop across the internal resistance is $\mathcal{E} - V_T = 1.0\ \text{V}$, so $r = \frac{1.0}{2.08} \approx 0.48\ \Omega$. This is choice C, the correct answer. Choice A is wrong because it uses only one branch current to find $r$, ignoring the current through $R_2$ entirely — this underestimates the total current and produces a meaningless "correction factor" that isn't physically justified. Choice B treats the parallel resistors as if they were in series, giving $R_{eq} = 10\ \Omega$. This is a fundamental circuit-topology error — parallel resistors always produce an equivalent resistance smaller than either individual resistor. Choice D claims the terminal voltage applies only to $R_1$, not $R_2$. This contradicts the definition of parallel circuits: all branches share the same voltage, which is exactly the terminal voltage. Your strategy: always identify the circuit topology first. In parallel circuits, voltage is shared — use $R_{eq}$ for the whole parallel network to find total current before solving for $r$.
7. As the battery ages and $r$ increases with $R$ and $\mathcal{E}$ held constant, which of the following correctly describes the simultaneous changes in terminal voltage $V_T$, current $I$, and power dissipated internally $P_{int}$?
   Answer: $V_T$ decreases, $I$ decreases, and $P_{int}$ may increase or decrease depending on whether $r < R$ or $r > R$. At $r = R$, $P_{int}$ reaches a maximum value of $\mathcal{E}^2/(4R)$, and for $r > R$, $P_{int}$ decreases back toward zero as $r \to \infty$. When a battery with fixed emf $\mathcal{E}$ drives current through a fixed external resistance $R$ plus an aging internal resistance $r$, every quantity you care about flows from one master equation: $I = \frac{\mathcal{E}}{R + r}$. Use this as your anchor. As $r$ increases, the denominator grows, so $I$ decreases — straightforward. Terminal voltage follows from $V_T = \mathcal{E} - Ir = \frac{\mathcal{E} R}{R+r}$, which also decreases as $r$ grows (larger $r$ in the denominator, smaller numerator contribution). So $V_T$ falls and $I$ falls — both together. Now for internal power: $P_{int} = I^2 r = \frac{\mathcal{E}^2 r}{(R+r)^2}$. This is not monotonic. Taking the derivative and setting it to zero reveals a maximum at $r = R$, giving $P_{int,max} = \frac{\mathcal{E}^2}{4R}$. Below $r = R$, increasing $r$ raises $P_{int}$; above $r = R$, the shrinking current wins and $P_{int}$ falls back toward zero. This is exactly what D describes — the correct answer. A is wrong because it correctly identifies the $r = R$ maximum but incorrectly states $V_T$ increases. The terminal voltage drops as $r$ grows, not recovers. B is wrong because it claims $P_{int}$ increases monotonically — ignoring that beyond $r = R$, the current collapse dominates and $P_{int}$ actually decreases. C is doubly wrong: $V_T$ decreases (not increases toward $\mathcal{E}$), and $P_{int}$ is non-monotonic, not uniformly falling. When you see competing effects — one quantity rising, another falling — always write out the explicit formula and check for a maximum analytically rather than reasoning qualitatively alone.
8. What is the terminal voltage of the battery during charging, and which expression correctly represents the power balance inside the battery?
   Answer: $V_T = 14 \text{ V}$; input power from charger $= V_T I = 56 \text{ W}$, power stored chemically $= \mathcal{E}I = 48 \text{ W}$, and power dissipated in $r$ $= I^2 r = 8 \text{ W}$, satisfying conservation of energy: $56 = 48 + 8$. When a battery is being charged, current is forced into it against its emf — this reverses the role of the internal resistance compared to discharging. Instead of the internal resistance reducing the terminal voltage below the emf, it now adds to it. The terminal voltage during charging is: $V_T = \mathcal{E} + Ir = 12 + (4)(0.5) = 14 \text{ V}$ The charger must push current through both the battery's emf and its internal resistance, so it supplies more voltage — and more power — than the emf alone. The power balance confirms this: input power $= V_T I = (14)(4) = 56 \text{ W}$, which splits into chemical storage $= \mathcal{E}I = (12)(4) = 48 \text{ W}$ and resistive loss $= I^2r = (16)(0.5) = 8 \text{ W}$. Since $48 + 8 = 56$, energy is perfectly conserved. This is answer B. Answer A gets the direction exactly backwards — it treats charging like discharging, subtracting $Ir$ instead of adding it, giving $V_T = 10 \text{ V}$. The power accounting then breaks down entirely. Answer C uses the right logic (adding $Ir$) but makes an arithmetic error: $I^2r = (16)(0.5) = 8 \text{ W}$, not $4 \text{ W}$, so its energy balance fails and the terminal voltage of 13 V is wrong. Answer D incorrectly claims internal resistance only matters during discharge. In reality, $Ir$ always creates a voltage drop across the internal resistance regardless of current direction. Study tip: Always ask yourself — is the battery charging or discharging? Discharging: $V_T = \mathcal{E} - Ir$. Charging: $V_T = \mathcal{E} + Ir$. Getting this direction right is the entire key to these problems.
9. Comparing the two configurations, which statement correctly describes the current delivered to $R$ and the terminal voltage across $R$ in each case?
   Answer: Series: $I_R = 1.8 \text{ A}$, $V_R = 14.4 \text{ V}$; Parallel: $I_R = 1.06 \text{ A}$, $V_R = 8.47 \text{ V}$. The parallel configuration delivers less current to $R$ but is preferable when longevity matters, since each battery supplies only half the total current, reducing internal losses per cell. When analyzing battery configurations, always start by finding the equivalent EMF and equivalent internal resistance for each setup — then apply Ohm's law to the full circuit. Series configuration: Two batteries in series add their EMFs and internal resistances: $\mathcal{E}_{total} = 18\text{ V}$, $r_{total} = 2\ \Omega$. The current through the circuit is $I = \frac{18}{8+2} = 1.8\text{ A}$. The terminal voltage across $R$ is $V_R = 1.8 \times 8 = 14.4\text{ V}$. Parallel configuration: Identical batteries in parallel keep the same EMF ($\mathcal{E} = 9\text{ V}$) but halve the internal resistance: $r_{eq} = 0.5\ \Omega$. The circuit current is $I = \frac{9}{8+0.5} \approx 1.06\text{ A}$, giving $V_R = 1.06 \times 8 \approx 8.47\text{ V}$. Each battery only supplies half this current ($\approx 0.53\text{ A}$), meaning lower internal dissipation per cell and longer battery life. This confirms B is correct. A has the right numbers but gives the wrong reasoning — it implies series is simply "better," ignoring the practical advantage of parallel for longevity. C incorrectly states $I = 2.0\text{ A}$ and $V_R = 16.0\text{ V}$, which would only be true if internal resistance were zero (it isn't — the $2\ \Omega$ drop is real). D claims the parallel combination has zero internal resistance, which is false; $r_{eq} = r/2 = 0.5\ \Omega$, not zero. Study tip: Memorize the two parallel-battery rules — same EMF, halved internal resistance — and always account for internal resistance in your voltage calculations; exams frequently trap students who forget that $V_R \neq \mathcal{E}_{total}$.

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