Physics 2 Quiz: Battery Emf And Internal Resistance

What this quiz covers

This quiz focuses on Battery Emf And Internal Resistance, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 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. (correct answer)
2. The y-intercept equals $\mathcal{E} - Ir$ evaluated at maximum load and the x-intercept equals $r$, because at zero current the voltage drop across the internal resistance vanishes and $r$ sets the scale of the current axis.
3. The y-intercept equals the terminal voltage under full load and the x-intercept equals $\mathcal{E}/R$, because the external resistance $R$ determines how much current flows at zero terminal voltage rather than the internal resistance $r$.
4. The y-intercept equals $\mathcal{E} + Ir$ and the x-intercept equals $r/\mathcal{E}$, because the terminal voltage rises above the emf when current flows due to back-emf effects inside the battery under charging conditions.

Question 2

1. $V_T = 10 \text{ V}$; power stored chemically $= \mathcal{E}I = 48 \text{ W}$, power dissipated in $r$ $= I^2 r = 8 \text{ W}$, and input power from charger $= V_T I = 40 \text{ W}$. Energy is conserved because the terminal voltage is lower than the emf during charging, so the charger supplies less power than what is stored.
2. $V_T = 14 \text{ V}$; 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: $56 = 48 + 8$. (correct answer)
3. $V_T = 13 \text{ V}$; input power from charger $= V_T I = 52 \text{ W}$, power stored chemically $= \mathcal{E}I = 48 \text{ W}$, and power dissipated in $r$ $= I^2 r = 4 \text{ W}$, satisfying conservation of energy: $52 = 48 + 4$. The terminal voltage exceeds the emf because the charger reverses the internal current direction, halving the resistive losses.
4. $V_T = 12 \text{ V}$; the terminal voltage equals the emf during charging because the internal resistance only drops voltage during discharge. The charger supplies exactly $\mathcal{E}I = 48 \text{ W}$, all of which is stored chemically with no resistive losses in the internal resistance.

Question 3

1. $r = 2.0 \text{ }\ \Omega$, found by noting that the voltage drop across $r$ is $1.0 \text{ V}$ and the current is estimated using only one branch: $I \approx V_T/R_1 = 5.0/4 = 1.25 \text{ A}$ (ignoring the second branch), giving $r = 1.0/1.25 = 0.80 \text{ }\ \Omega$, but then corrected upward to $2.0 \text{ }\ \Omega$ by the parallel branch factor.
2. $r = 0.60 \text{ }\ \Omega$, found by adding $R_1$ and $R_2$ in series to get $R_{eq} = 10 \text{ }\ \Omega$, computing $I = \mathcal{E}/(R_{eq} + r)$ and solving with $V_T = \mathcal{E} - Ir$, which gives $r = \mathcal{E} R_{eq}(\mathcal{E} - V_T)^{-1} - R_{eq}$.
3. $r = 0.48 \text{ }\ \Omega$, 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 $r = (\mathcal{E} - V_T)/I = 1.0/2.08 \approx 0.48 \text{ }\ \Omega$. (correct answer)
4. $r = 0.48 \text{ }\ \Omega$, but this result assumes the terminal voltage equals the voltage across $R_1$ only; the voltage across $R_2$ is different because the resistors are in parallel, requiring separate current calculations for each branch before summing.

Question 4

1. $I = 2.5 \text{ A}$ and battery 2 is being charged, because the opposing polarity means the net emf is $\mathcal{E}_1 - \mathcal{E}_2 = 4 \text{ V}$, and current flows in the direction driven by the stronger battery (battery 1), which forces current backward through battery 2, charging it.
2. $I = 0.50 \text{ A}$ and battery 1 is being charged, because the net emf in the loop is $\mathcal{E}_1 - \mathcal{E}_2 = 4 \text{ V}$, total resistance is $r_1 + r_2 + R = 8 \text{ }\ \Omega$, and the current flows in the direction set by battery 2 (the larger emf determines direction), which pushes current backward through battery 1, charging it.
3. $I = 2.5 \text{ A}$ and battery 1 is being charged, because the total emf is $\mathcal{E}_1 + \mathcal{E}_2 = 20 \text{ V}$ (opposing polarities add when batteries face each other), total resistance is $r_1 + r_2 + R = 8 \text{ }\ \Omega$, but only $R = 5 \text{ }\ \Omega$ limits the current, giving $I = 20/8 = 2.5 \text{ A}$.
4. $I = 0.50 \text{ A}$ 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. (correct answer)

Question 5

1. Efficiency is exactly 75%, because at $R = r$ the power delivered to $R$ is $\mathcal{E}^2/(4r)$ and the total power from the emf is $\mathcal{E}^2/(2r)$, so the fraction delivered to $R$ is $1/2 = 50\%$, and the remaining 25% accounts for impedance matching overhead that reduces apparent losses.
2. 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. (correct answer)
3. Efficiency is exactly 50%, but only when the battery is fully charged; as the battery ages and $r$ increases while $\mathcal{E}$ decreases, the optimal $R$ shifts upward while efficiency rises above 50%, meaning older batteries are paradoxically more efficient at maximum power transfer.
4. Efficiency approaches 100% at maximum power transfer because the condition $R = r$ minimizes the voltage drop across the internal resistance, and as $R$ is tuned to match $r$, nearly all the emf appears across the external load with negligible internal dissipation.

Question 6

1. $r = 1.25 \text{ } \Omega$ 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 $r = 1.0/0.80 = 1.25 \text{ } \Omega$. (correct answer)
2. $r = 1.11 \text{ } \Omega$ and $I = 0.90 \text{ A}$, found by treating $V_1 = 9.0 \text{ V}$ as the terminal voltage under load rather than the open-circuit emf, computing current as $I = V_1/R = 9.0/10 = 0.90 \text{ A}$, and then using $r = (V_1 - V_2)/I = 1.0/0.90 \approx 1.11 \text{ } \Omega$.
3. $r = 0.125 \text{ } \Omega$ and $I = 0.80 \text{ A}$, found by computing the current correctly as $I = V_2/R = 0.80 \text{ A}$ but using the loaded terminal voltage $V_2 = 8.0 \text{ V}$ as the emf, so $r = (V_2 - V_2)/I = 0$, then approximating from the ratio $\Delta V/V_1 = 1/9 \approx 0.111$ times $R$ to get $r \approx 0.125 \text{ } \Omega$.
4. $r = 1.25 \text{ } \Omega$ and $I = 0.80 \text{ A}$, but the emf cannot be determined from these two measurements alone because $V_1$ was measured with a non-ideal voltmeter that draws a small current, so the true open-circuit voltage is slightly higher than 9.0 V and additional information about the voltmeter resistance is needed.

Question 7

1. $\mathcal{E} = 12.15 \text{ V}$ and $r = 0.25 \text{ } \Omega$, but this result is only valid if the battery's emf remains constant between discharge and charge. In practice, $\mathcal{E}$ shifts with state of charge, so the true internal resistance requires a correction factor and cannot be determined from just two measurements at a single current.
2. $\mathcal{E} = 12.15 \text{ V}$ and $r = 0.50 \text{ } \Omega$, obtained by applying $V_T = \mathcal{E} - Ir$ to both measurements (treating both as discharge): $11.4 = \mathcal{E} - 3r$ and $12.9 = \mathcal{E} - 3r$. Since these two equations are inconsistent, the student averages the right-hand sides to get $\mathcal{E} = 12.15 \text{ V}$ and estimates $r = (12.9 - 11.4)/(2 \times 3) = 0.25 \text{ } \Omega$, then doubles $r$ to $0.50 \text{ } \Omega$ to account for the sign reversal during charging.
3. $\mathcal{E} = 12.15 \text{ V}$ 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 $r = 1.5/6 = 0.25 \text{ } \Omega$). (correct answer)
4. $\mathcal{E} = 11.4 \text{ V}$ and $r = 0.50 \text{ } \Omega$, obtained by identifying the discharge terminal voltage as the emf (since the battery acts as a source during discharge), then using the difference between the two readings: $r = (12.9 - 11.4)/(2 \times 3) = 0.25 \text{ } \Omega$, doubled to $0.50 \text{ } \Omega$ because current flows in opposite directions in the two scenarios.

Question 8

1. 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 series configuration delivers more current and higher terminal voltage to $R$ because the emfs add while the internal resistances also add, yielding a net gain in driving voltage.
2. 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. (correct answer)
3. Series: $I_R = 2.0 \text{ A}$, $V_R = 16.0 \text{ V}$; Parallel: $I_R = 1.06 \text{ A}$, $V_R = 8.47 \text{ V}$. The series configuration doubles both the emf and the terminal voltage because internal resistance is negligible compared to $R$, so the full $18 \text{ V}$ appears across $R$.
4. Series: $I_R = 1.8 \text{ A}$, $V_R = 14.4 \text{ V}$; Parallel: $I_R = 1.13 \text{ A}$, $V_R = 9.0 \text{ V}$. The parallel configuration maintains the full $9 \text{ V}$ emf across $R$ with no internal resistance losses, because the two internal resistances cancel when placed in parallel, making the combination an ideal voltage source.

Question 9

1. $V_T$ increases, $I$ decreases, and $P_{int}$ may increase or decrease depending on whether $r < R$ or $r > R$. At $r = R$, $P_{int}$ is maximized (by the maximum power transfer theorem applied to the internal resistance), and beyond this point $P_{int}$ decreases even as $r$ continues to rise.
2. $V_T$ decreases, $I$ decreases, and $P_{int}$ increases monotonically as $r$ grows, because the larger internal resistance always dissipates more power regardless of the ratio $r/R$, since $P_{int} = I^2 r$ and $r$ is increasing even as $I$ falls.
3. $V_T$ increases, $I$ decreases, and $P_{int}$ decreases monotonically as $r$ grows, because the terminal voltage recovering toward $\mathcal{E}$ means less voltage is dropped internally, so both the current and the internal dissipation fall together as the battery approaches open-circuit conditions.
4. $V_T$ 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 $r \to \infty$. (correct answer)