Ohm's Law & Power - Physics 2

1. What is the power delivered to the parallel combination of resistors?
   Answer: $32\,\text{W}$, because the equivalent parallel resistance is $8\,\Omega$, the total circuit resistance is $10\,\Omega$, the current is $I = 20/10 = 2\,\text{A}$, and $P_{\text{ext}} = I^2 R_{\text{ext}} = (2)^2(8) = 32\,\text{W}$. When a battery with internal resistance powers an external circuit, the internal resistance acts like a resistor in series with everything else — it "steals" some of the voltage before the current even reaches the external components. Your job is to find how much voltage (and therefore power) actually reaches the parallel combination. Start by finding the equivalent resistance of the parallel combination: $\frac{1}{R_{\text{ext}}} = \frac{1}{10} + \frac{1}{40} = \frac{5}{40}$, so $R_{\text{ext}} = 8\,\Omega$. The total circuit resistance is $R_{\text{total}} = r + R_{\text{ext}} = 2 + 8 = 10\,\Omega$. Ohm's law gives the current: $I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{20}{10} = 2\,\text{A}$. The power delivered to the parallel combination is then $P_{\text{ext}} = I^2 R_{\text{ext}} = (2)^2(8) = 32\,\text{W}$, confirming D. Choice A reaches the right terminal voltage ($16\,\text{V}$) but then misapplies it — mixing up total current with branch current in a contradictory way, producing a meaningless hybrid calculation. Choice B correctly finds total power from the source ($\mathcal{E} \cdot I = 40\,\text{W}$) but ignores the fact that some power is dissipated internally — the internal resistance always takes its share, regardless of how small it seems. Choice C assumes the terminal voltage equals the full EMF, which would only be true if $r = 0$; with $r = 2\,\Omega$ carrying $2\,\text{A}$, there's a real $4\,\text{V}$ drop across the internal resistance. As a rule, always account for internal resistance by finding total resistance first, then solving for current — never assume the terminal voltage equals the EMF unless explicitly told the battery is ideal.
2. What value of $R_3$ must the student use, and which of the following correctly explains why a common intuitive approach leads to the wrong answer?
   Answer: $R_3 = 2\,\Omega$. A common wrong approach is to average the two resistances, giving $(R_1 + R_2)/2 = 4.5\,\Omega$; this underestimates the correct equivalent resistance by more than a factor of 2 and would predict only $144/4.5 \approx 32\,\text{W}$, less than half the correct power of $72\,\text{W}$. Whenever you see resistors in parallel, your first instinct should be to find the equivalent resistance using the reciprocal formula, then use that to analyze power — not to average or add the resistances directly. For two resistors in parallel, the equivalent resistance is $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$, giving $R_{eq} = 2\,\Omega$. The total power delivered by the source is $P = \frac{V^2}{R_{eq}} = \frac{144}{2} = 72\,\text{W}$. To preserve this power, $R_3$ must equal $2\,\Omega$ — confirming answer C. Notice that $2\,\Omega$ is strictly less than the smaller resistor ($3\,\Omega$), which is always true for parallel combinations. Choice C also correctly identifies the intuitive trap: averaging the resistances gives $4.5\,\Omega$, which would predict only $144/4.5 \approx 32\,\text{W}$ — less than half the actual power. Choice A describes a student who adds the resistances as if they were in series ($9\,\Omega$), a classic series/parallel mix-up. This overestimates resistance dramatically and delivers only $16\,\text{W}$. Choice B invents a fictional two-step averaging-then-doubling procedure that has no grounding in circuit theory and also lands on $9\,\Omega$. Choice D gets the resistance value wrong ($4.5\,\Omega$ instead of $2\,\Omega$) and merely restates the rule that parallel resistance is less than the smallest branch — it identifies a true principle but applies it to the wrong number. Study tip: For parallel resistors, always use the reciprocal formula. A quick sanity check — your answer must be smaller than the smallest individual resistor — will catch most errors immediately.
3. What is the ratio of the power dissipated in $R_C$ to the power dissipated in $R_A$?
   Answer: $P_C : P_A = 4 : 1$, because $R_C$ carries the full circuit current $I$ while $R_A$ carries only $I/2$, and since $P \propto I^2R$ with equal resistances, the ratio is $(I)^2 R \,:\, (I/2)^2 R = 4 : 1$. Whenever you see a resistor network problem asking about power ratios, your first move should be to find the current through each element — then apply $P = I^2 R$ carefully, remembering that power scales with the square of current, not linearly. Here's the circuit analysis: $R_A$ and $R_B$ are in parallel, giving a combined resistance of $R/2$. That combination is in series with $R_C = R$, so the total resistance is $3R/2$. The full circuit current is $I = \mathcal{E}/(3R/2) = 2\mathcal{E}/3R$. This entire current $I$ flows through $R_C$. At the parallel junction, it splits equally, so each of $R_A$ and $R_B$ carries only $I/2$. Since all resistances are equal, the power ratio is: $\frac{P_C}{P_A} = \frac{I^2 R}{(I/2)^2 R} = \frac{I^2}{I^2/4} = 4$ So $P_C : P_A = 4 : 1$, confirming C is correct. A is the classic trap of treating power as proportional to current rather than current-squared. $R_C$ carries twice the current of $R_A$, but that means four times the power, not two. B is wrong because symmetry only applies to $R_A$ and $R_B$ (which carry equal currents), not to $R_C$, which sits in a fundamentally different position in the circuit. D inverts the correct ratio entirely — higher voltage across $R_C$ means more power, not less. Study tip: Always write down $P = I^2 R$ explicitly before comparing power. The squared relationship is the most commonly forgotten detail in circuit power problems.
4. What is the mechanical power output of the motor?
   Answer: $550\,\text{W}$, because the total input power is $P_{\text{in}} = VI = 600\,\text{W}$, and the power lost to heat in the coil resistance is $P_R = I^2R = 50\,\text{W}$, leaving $550\,\text{W}$ for mechanical output. When a real electric motor operates, it behaves as both a resistor (the coil winding) and a generator (the spinning rotor producing a back-EMF). The total voltage from the supply must drive current through both of these effects simultaneously. Your job is to separate the power budget into "wasted heat" and "useful mechanical work." The total electrical power delivered by the source is $P_{\text{in}} = VI = (120)(5) = 600\,\text{W}$. Not all of this becomes mechanical output — some is dissipated as heat in the coil's resistance. That resistive loss is $P_R = I^2R = (5)^2(2) = 50\,\text{W}$. By conservation of energy, the mechanical power output is simply what remains: $P_{\text{mech}} = 600 - 50 = 550\,\text{W}$. That's answer A, and it's correct. Choice B is wrong because it assumes a perfectly ideal motor with zero resistance — but the problem explicitly gives you a $2\,\Omega$ coil, which always dissipates heat. Choice C contains a seductive trap: it correctly identifies the back-EMF as $\mathcal{E}_{\text{back}} = V - IR = 110\,\text{V}$ and correctly computes $\mathcal{E}_{\text{back}} \cdot I = 550\,\text{W}$, but then wrongly subtracts the 50 W heat loss a second time — that loss is already excluded when you use back-EMF times current, since the back-EMF method and the energy-subtraction method are two equivalent routes to the same answer of 550 W, not steps to chain together. Choice D confuses cause and effect entirely: resistive heating is wasted energy, not useful mechanical work. A reliable strategy: for motor problems, always start with $P_{\text{in}} = VI$, then subtract $I^2R$ for heat loss. The remainder is mechanical power — clean, simple, and hard to mess up.
5. By approximately what percentage does the power dissipated in the heating element decrease during the brownout?
   Answer: Approximately 19%, because power is proportional to $V^2$, and a 10% decrease in voltage gives $(0.9)^2 = 0.81$, representing a 19% decrease in power. Whenever you see a question involving power dissipation with a fixed resistance, your instinct should be to reach for $P = \frac{V^2}{R}$. This formula reveals a critical insight: power depends on the square of voltage, not voltage directly. Here's why B is correct. If voltage drops by 10%, the new voltage is $0.9V$. Plugging into the formula: $P_{new} = \frac{(0.9V)^2}{R} = \frac{0.81V^2}{R} = 0.81 \cdot P_{original}$. The power becomes 81% of its original value — a 19% decrease. You can verify with numbers: the original resistance is $R = \frac{V^2}{P} = \frac{(120)^2}{1200} = 12\,\Omega$. At 108 V, $P = \frac{(108)^2}{12} = \frac{11664}{12} = 972\,\text{W}$, which is indeed about 19% less than 1200 W. A is the most common trap — it assumes a linear relationship between voltage and power. That's only true if current were held constant (like in $P = IV$ with fixed $I$), but here resistance is fixed, not current. C correctly identifies that both $I$ and $V$ drop by 10%, and that $P = IV$ gives a combined effect — but the math is $(0.9)(0.9) = 0.81$, which is exactly 19%, not "approximately 20% corrected to 19%." The framing is unnecessarily confusing and circular. D introduces a real-world nuance (resistance decreasing as the element cools), but the problem explicitly states resistance remains constant, so this consideration is ruled out by the given assumptions. Your study tip: memorize that for fixed resistance, $P \propto V^2$ — a 10% voltage drop means roughly a 19% power drop, not 10%. This squared relationship appears frequently on Physics 2 exams.
6. Which statement best characterizes the device, and what is the power dissipated at $V = 4\,\text{V}$?
   Answer: The device is non-ohmic because the ratio $V/I$ is not constant; $I \propto V^2$ fits the data, and the power at $4\,\text{V}$ is $8\,\text{W}$, calculated directly as $P = IV = (2)(4)$. When a question gives you multiple data points for a device, your first move should always be to check whether $V/I$ is constant — that's the test for ohmic behavior. Here, the ratios are $2/0.5 = 4\,\Omega$, $4/2 = 2\,\Omega$, and $6/4.5 \approx 1.33\,\Omega$. The ratio is clearly not constant, so the device is non-ohmic. Next, look for a pattern: notice that when voltage doubles (2 V → 4 V), current quadruples (0.5 A → 2 A). That's the signature of $I \propto V^2$. You can verify: if $I = kV^2$, then $0.5 = k(4)$ gives $k = 0.125$, and indeed $0.125 \times 16 = 2\,\text{A}$ and $0.125 \times 36 = 4.5\,\text{A}$. Power at 4 V is simply $P = IV = (2\,\text{A})(4\,\text{V}) = 8\,\text{W}$. That's choice B, the correct answer. Choice A is wrong on two counts: $V/I$ is not constant (it changes at every point), and even the first data point gives $R = 4\,\Omega$, not a valid fixed resistance for the device. C correctly identifies the non-ohmic behavior and the $I \propto V^2$ relationship, but then misapplies $P = I^2/R$ using a fabricated "effective resistance" of $2\,\Omega$ — this formula only works for true ohmic devices with a fixed $R$. D misreads the effective resistance at 4 V; $R_{\text{eff}} = V/I = 4/2 = 2\,\Omega$, not $1\,\Omega$, so the power calculation is wrong. The universal power formula $P = IV$ always works regardless of device type — memorize it as your safe default whenever the device behavior is unknown or non-ohmic.
7. For the output voltage across $R_L$ to equal exactly $9\,\text{V}$, what must $R_L$ equal, and what is the power dissipated in $R_1$ under this condition?
   Answer: $R_L$ must be infinite (open circuit), and the power in $R_1$ is $9\,\text{mW}$, because only with no load current can the unloaded divider ratio $R_2/(R_1+R_2) = 3/4$ yield $9\,\text{V}$, with current $I = 12/4\,\text{k}\Omega = 3\,\text{mA}$ and $P_{R_1} = I^2 R_1 = (3\,\text{mA})^2(1\,\text{k}\Omega) = 9\,\text{mW}$. Voltage divider questions test your ability to analyze how a parallel load disturbs an existing series circuit — and recognize when "no disturbance" is itself the answer. An unloaded voltage divider splits the supply proportionally: $V_{out} = V_s \cdot \frac{R_2}{R_1 + R_2}$. Plugging in: $V_{out} = 12 \cdot \frac{3}{1+3} = 9\,\text{V}$. The divider already delivers exactly 9 V across $R_2$ with no load attached. The moment you connect any finite $R_L$ in parallel with $R_2$, the effective resistance drops below $3\,\text{k}\Omega$, pulling the output voltage below 9 V. Therefore, the only way to maintain exactly 9 V is to leave $R_L$ as an open circuit (infinite resistance). With that, the single loop carries $I = 12\,\text{V} / 4\,\text{k}\Omega = 3\,\text{mA}$, and the power in $R_1$ is $P = I^2 R_1 = (3\times10^{-3})^2 \times 1000 = 9\,\text{mW}$. That confirms D. A is self-contradictory — it correctly computes that $9\,\text{k}\Omega \parallel 3\,\text{k}\Omega = 2.25\,\text{k}\Omega$ gives only 8.3 V, then invents a nonsensical "compensation" mechanism rather than accepting the result disproves the premise. B makes the same structural error: it connects a finite load, correctly finds the output drops to 7.2 V, then wrongly suggests adjusting $R_L$ upward fixes things — which only works in the infinite limit (answer D). C reaches the correct $R_L = \infty$ conclusion but fabricates the idea that energy is "stored in the electric field of $R_2$" and adds it to the resistor's dissipation. Resistors dissipate power — they don't store it. Strategy tip: When a voltage divider question asks what load produces a specific output, always check the unloaded ratio first. If it already matches, the answer is no load at all — a classic trap that rewards students who calculate before assuming a finite $R_L$ is required.
8. What fraction of the total power delivered by the battery is dissipated as heat in the internal resistance of the battery?
   Answer: $\dfrac{1}{7}$, because the internal resistance is $\frac{1}{7}$ of the total resistance in the circuit, and power dissipated is proportional to resistance when current is the same through all elements. When a circuit has resistors in series — including a battery's internal resistance — the same current flows through every element. That single current is the key to finding power ratios. First, find the total resistance: $R_{total} = r + R_1 + R_2 = 2 + 4 + 8 = 14\,\Omega$. The current is $I = \mathcal{E}/R_{total} = 28/14 = 2\,\text{A}$. Since power dissipated in any resistor is $P = I^2 R$, and I is identical for every element, the fraction of total power lost in any resistor is simply that resistor's share of the total resistance. For the internal resistance: $\frac{P_r}{P_{total}} = \frac{I^2 r}{I^2 R_{total}} = \frac{r}{R_{total}} = \frac{2}{14} = \frac{1}{7}$. Answer A is correct. Answer B claims the power ratio equals the voltage ratio, but that's a double error — the voltage ratio across internal resistance is $2/28 = 1/14$, and even if you used it, power ratios in series circuits don't equal voltage ratios (power scales as $V^2/R$, not $V$). Answer C incorrectly compares $r$ only to $R_2 = 8\,\Omega$ rather than to the full circuit resistance of $14\,\Omega$, ignoring $R_1$ entirely. Answer D is pure fabrication — in a series circuit, the current through every element is identical, so there's no "doubling" of current through the internal resistance. Study tip: In any series circuit, when current is shared, power fractions reduce to simple resistance ratios. Memorize: $P_x/P_{total} = R_x/R_{total}$ for series elements.
9. A cylindrical resistor of length $L$, cross-sectional area $A$, and resistivity $\rho$ carries a current $I$. The resistor is then replaced by one made of the same material but with twice the length and twice the radius. By what factor does the power dissipated in the resistor change if the voltage across it is held constant?
   Answer: The power increases by a factor of 2, because the new resistance is $R' = \rho(2L)/[\pi(2r)^2] = \rho(2L)/(4\pi r^2) = R/2$, so with constant voltage $P = V^2/R$ gives $P' = 2V^2/R = 2P$. Whenever you see a resistor problem involving geometry changes, your first move should always be to find the new resistance using $R = \rho L / A$, then apply the correct power formula based on what's held constant — voltage or current. That choice of formula is everything. Here, the new resistor has twice the length ($2L$) and twice the radius ($2r$), so its cross-sectional area becomes $A' = \pi(2r)^2 = 4\pi r^2 = 4A$. Plugging into the resistance formula: $R' = \rho(2L)/(4A) = \frac{1}{2} \cdot \frac{\rho L}{A} = R/2$. The resistance halves. Since voltage is held constant, you use $P = V^2/R$. With resistance cut in half, power doubles: $P' = V^2/(R/2) = 2V^2/R = 2P$. Answer B is correct. Answer A makes a critical geometry error — it acknowledges that length doubles but ignores that the area also quadruples, so the net effect on resistance is actually a decrease, not an increase. Answer C correctly finds $R' = R/2$, but then uses $P = I^2 R$ with the original current. That formula is only valid when current is held constant. Since voltage is held constant here, current changes too, making that approach invalid. Answer D is a tempting trap — the ratio $L/r$ is indeed preserved (both double), but resistance depends on $L/A = L/(\pi r^2)$, not $L/r$. Doubling both $L$ and $r$ does not preserve resistance. Study tip: Always identify what's held constant (voltage or current) before choosing your power formula — $P = V^2/R$ vs. $P = I^2R$ give opposite trends when resistance changes, and that's a classic exam trap.

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