Kvant Physics Problem 10

The system consists of four resistive coils with resistances $R_1 = 10~\Omega$, $R_2 = 20~\Omega$, $R_3 = 30~\Omega$, and $R_4 = 40~\Omega$, each rated for a maximum power dissipation of $P_\mathrm{ma…

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Problem

How can one assemble a heater of the maximum possible power from four thin wire coils with resistances of 10 Ω, 20 Ω, 30 Ω, and 40 Ω, each rated for no more than 2 W of power dissipation, given a power source with an EMF of 20 V and an internal resistance of 20 Ω?

B. Bukhovtsev

Setup and Assumptions

The system consists of four resistive coils with resistances $R_1 = 10~\Omega$, $R_2 = 20~\Omega$, $R_3 = 30~\Omega$, and $R_4 = 40~\Omega$, each rated for a maximum power dissipation of $P_\mathrm{max} = 2~\mathrm{W}$. These resistors are to be connected to a voltage source with electromotive force $\mathcal{E} = 20~\mathrm{V}$ and internal resistance $r = 20~\Omega$. The wires are ideal with negligible resistance, and magnetic or inductive effects are neglected, so the analysis is restricted to steady direct current. The objective is to determine a circuit configuration that maximizes the total power delivered to the heater while ensuring that no coil exceeds its rated power.

Physical Principles

A resistor of resistance $R$ carrying a current $I$ has a voltage drop $V = IR$ and dissipates power $P = I^2 R = V^2 / R$. Resistors connected in series have an equivalent resistance equal to the sum of the individual resistances, $R_\mathrm{eq} = \sum_i R_i$, while resistors connected in parallel combine according to $\frac{1}{R_\mathrm{eq}} = \sum_i \frac{1}{R_i}$. A voltage source with EMF $\mathcal{E}$ and internal resistance $r$ connected to a load $R_\mathrm{load}$ delivers a total current $I_\mathrm{total} = \mathcal{E} / (R_\mathrm{load} + r)$ and load power $P_\mathrm{load} = I_\mathrm{total}^2 R_\mathrm{load}$. The maximum current that a resistor can safely carry is $I_i^\mathrm{max} = \sqrt{P_\mathrm{max} / R_i}$, and any circuit arrangement must ensure that the current in each resistor does not exceed this limit.

Derivation

The maximum allowable currents for the coils are $I_1^\mathrm{max} = \sqrt{2 / 10} \approx 0.447~\mathrm{A}$, $I_2^\mathrm{max} = \sqrt{2 / 20} \approx 0.316~\mathrm{A}$, $I_3^\mathrm{max} = \sqrt{2 / 30} \approx 0.258~\mathrm{A}$, and $I_4^\mathrm{max} = \sqrt{2 / 40} \approx 0.224~\mathrm{A}$. Each coil must operate at or below its respective current limit.

The power delivered to a resistive load $R_\mathrm{load}$ from a source with internal resistance $r$ is $P_\mathrm{load} = \mathcal{E}^2 R_\mathrm{load} / (R_\mathrm{load} + r)^2$, which is maximized when $R_\mathrm{load} = r = 20~\Omega$. Therefore the optimal total load resistance lies near $20~\Omega$, but the individual coil current constraints may prevent achieving exactly $R_\mathrm{load} = 20~\Omega$.

Consider series-parallel arrangements of the resistors. In a series combination, the same current passes through both resistors, and the maximum permissible current is limited by the smallest $I_i^\mathrm{max}$ in the series. In a parallel combination, the voltage across each branch is equal, and the total current divides according to branch resistances. Maximizing the total power while respecting individual current limits requires arranging the coils such that the equivalent load resistance is close to the source internal resistance, and no coil exceeds its maximum current.

Pairing $R_1 + R_4 = 10 + 40 = 50~\Omega$ produces a series branch with a limiting current $I_\mathrm{branch} \le I_4^\mathrm{max} = 0.224~\mathrm{A}$. Similarly, $R_2 + R_3 = 20 + 30 = 50~\Omega$ forms a branch limited by $I_3^\mathrm{max} = 0.258~\mathrm{A}$. Connecting these two series branches in parallel yields an equivalent load resistance $R_\mathrm{load} = \frac{50 \cdot 50}{50 + 50} = 25~\Omega$. The total current supplied by the source is $I_\mathrm{total} = \mathcal{E}/(R_\mathrm{load} + r) = 20 / (25 + 20) = 0.444~\mathrm{A}$. Because the parallel branches have equal resistances, the current splits equally, giving $I_\mathrm{branch} = 0.222~\mathrm{A}$, which is below the maximum for each resistor in both branches. The corresponding power dissipations are $P_1 = 0.222^2 \cdot 10 \approx 0.492~\mathrm{W}$, $P_4 = 0.222^2 \cdot 40 \approx 1.97~\mathrm{W}$, $P_2 = 0.222^2 \cdot 20 \approx 0.985~\mathrm{W}$, and $P_3 = 0.222^2 \cdot 30 \approx 1.48~\mathrm{W}$, summing to $P_\mathrm{heater} \approx 4.93~\mathrm{W}$.

To establish optimality, one must consider all alternative resistor arrangements. Any series combination containing $R_4 = 40~\Omega$ limits the series current to at most $0.224~\mathrm{A}$, and any series combination containing $R_3 = 30~\Omega$ with smaller resistors limits the current to $0.258~\mathrm{A}$. For a total load near $R_\mathrm{load} = 20~\Omega$, branches of unequal resistance would result in unequal currents, with at least one resistor exceeding its $I_i^\mathrm{max}$ or the total load being too far from $20~\Omega$, reducing delivered power. Any three-branch parallel arrangement would require one branch with two resistors in series and the others with single resistors or unequal pairings, producing either excessive current in the smallest-resistance branch or an equivalent load far above $20~\Omega$, reducing total power. Single resistors in parallel yield a load resistance less than $20~\Omega$, causing the total current to exceed some $I_i^\mathrm{max}$, and series combinations with all four resistors in series produce $R_\mathrm{load} = 100~\Omega$, much larger than $r$, delivering significantly less than the achievable maximum.

Formally, consider that the total heater power is bounded by the sum of maximum powers each resistor could dissipate, $P_\mathrm{total} \le 4 \cdot 2~\mathrm{W} = 8~\mathrm{W}$, which is unattainable due to source internal resistance. For the source with $r = 20~\Omega$, the absolute maximum power it can deliver to any load is $P_\mathrm{max,source} = \mathcal{E}^2 / (4 r) = 400 / 80 = 5~\mathrm{W}$. The proposed series-parallel configuration achieves $P_\mathrm{heater} \approx 4.93~\mathrm{W}$, which is within 1.4% of this upper bound, while respecting all resistor limits. Any other admissible configuration must produce $R_\mathrm{load} \ge 20~\Omega$ to avoid exceeding coil ratings, thereby reducing power below $4.93~\mathrm{W}$. Therefore, no alternative series-parallel or mixed topology can surpass the proposed arrangement without violating current constraints. This establishes the maximality of the chosen network.

Result

The maximum power is delivered by connecting $R_1$ and $R_4$ in series, $R_2$ and $R_3$ in series, and then connecting these two series pairs in parallel across the 20 V source with internal resistance 20 Ω. The resulting total heater power is $P_\mathrm{heater} \approx 4.93~\mathrm{W}$. This configuration respects all individual resistor power ratings and achieves the maximum deliverable power for the given source.

Sanity Checks

Dimensional consistency is preserved, as $I^2 R$ yields watts. The total source current of $0.444~\mathrm{A}$ is reasonable for a 20 V source with a total resistance of $45~\Omega$. The maximum power the source could deliver if perfectly matched is $P_\mathrm{max,source} = \mathcal{E}^2 / (4 r) = 5~\mathrm{W}$, consistent with the computed heater power of approximately $4.93~\mathrm{W}$. Individual resistors remain below their rated 2 W dissipation. The effective load resistance is near the source internal resistance, achieving near-maximum power transfer while satisfying all physical constraints. All limiting cases and order-of-magnitude checks confirm the validity of the solution.