Kvant Physics Problem 10

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m19s
Source on kvant.digital

Problem

How can four thin wire coils with resistances of 10 ohms, 20 ohms, 30 ohms, and 40 ohms, each designed to dissipate no more than 2 W of power, be used to construct a heater of maximum possible power if there is a current source with an emf of 20 V and an internal resistance of 20 ohms?

B. Bukhovtsev

Setup and Assumptions

The system consists of a real voltage source with electromotive force $\mathcal{E} = 20,\text{V}$ and internal resistance $r = 20,\Omega$, connected to an external heater assembled from four thin wire coils with resistances $R_1 = 10,\Omega$, $R_2 = 20,\Omega$, $R_3 = 30,\Omega$, and $R_4 = 40,\Omega$. Each coil is thermally limited so that the power dissipated in it satisfies $P_i \le 2,\text{W}$.

The task is to choose a series and parallel connection of the four coils forming an equivalent load resistance $R_{\text{ext}}$ that maximizes the total thermal power in the coils. All wires are assumed ohmic with temperature-independent resistance, connections are ideal, and the internal resistance of the source is concentrated in $r$. Magnetic effects and inductance are neglected because the process is steady state DC.

The unknown is the optimal circuit configuration and the corresponding maximum total power delivered to the heater.

Physical Principles

The current delivered by a source with internal resistance is

$$I = \frac{\mathcal{E}}{r + R_{\text{ext}}}.$$

The power dissipated in the external load is

$$P_{\text{ext}} = I^2 R_{\text{ext}} = \frac{\mathcal{E}^2 R_{\text{ext}}}{(r + R_{\text{ext}})^2}.$$

For any individual resistor carrying current $I_k$, the Joule heating is

$$P_k = I_k^2 R_k.$$

In series elements the current is identical, while in parallel branches the voltage is identical across branches. The physical constraint requires $P_k \le 2,\text{W}$ for every coil.

The total useful power is maximized when $R_{\text{ext}}$ is as close as possible to the value that maximizes $P_{\text{ext}}$ under unconstrained conditions, namely $R_{\text{ext}} = r$, while still satisfying all individual element power limits.

Derivation

The external power as a function of equivalent resistance is

$$P_{\text{ext}}(R_{\text{ext}}) = \frac{(20,\text{V})^2 R_{\text{ext}}}{(20,\Omega + R_{\text{ext}})^2}.$$

Differentiation with respect to $R_{\text{ext}}$ gives the extremum condition

$$\frac{d}{dR_{\text{ext}}}\left(\frac{R_{\text{ext}}}{(r + R_{\text{ext}})^2}\right)=0,$$

which yields $R_{\text{ext}} = r = 20,\Omega$. The unconstrained optimum therefore corresponds to matching the external resistance to $20,\Omega$.

A direct construction of exactly $20,\Omega$ using $10,\Omega$, $20,\Omega$, $30,\Omega$, and $40,\Omega$ while respecting individual power limits is not feasible because any configuration producing $20,\Omega$ requires strongly uneven current distribution that forces at least one coil above $2,\text{W}$ under the resulting operating current. The optimal strategy is therefore to search for a realizable equivalent resistance close to $20,\Omega$ that satisfies all constraints.

A symmetric and physically balanced arrangement is obtained by forming two series pairs:

one branch consists of $10,\Omega$ in series with $40,\Omega$, giving $50,\Omega$, and the second branch consists of $20,\Omega$ in series with $30,\Omega$, also giving $50,\Omega$. These two identical branches are connected in parallel, so the external resistance is

$$R_{\text{ext}} = \frac{50,\Omega \cdot 50,\Omega}{50,\Omega + 50,\Omega} = 25,\Omega.$$

The total current from the source is

$$I = \frac{20,\text{V}}{20,\Omega + 25,\Omega} = \frac{20}{45},\text{A} \approx 0.444,\text{A}.$$

The voltage across the heater is

$$U = I R_{\text{ext}} = \frac{20}{45}\cdot 25,\text{V} \approx 11.11,\text{V}.$$

Since the branches are identical, the current splits equally:

$$I_{\text{branch}} = \frac{U}{50,\Omega} \approx \frac{11.11}{50},\text{A} \approx 0.222,\text{A}.$$

Power in each resistor is computed from $P = I^2 R$. In the $10,\Omega$ resistor,

$$P_{10} = (0.222,\text{A})^2 \cdot 10,\Omega \approx 0.494,\text{W}.$$

In the $40,\Omega$ resistor,

$$P_{40} = (0.222,\text{A})^2 \cdot 40,\Omega \approx 1.98,\text{W}.$$

In the other branch,

$$P_{20} \approx (0.222,\text{A})^2 \cdot 20,\Omega \approx 0.99,\text{W},$$

$$P_{30} \approx (0.222,\text{A})^2 \cdot 30,\Omega \approx 1.48,\text{W}.$$

All values remain below the allowed limit of $2,\text{W}$, so the configuration is physically admissible.

The total heater power equals

$$P_{\text{ext}} = I^2 R_{\text{ext}} = (0.444,\text{A})^2 \cdot 25,\Omega \approx 4.94,\text{W}.$$

Result

The optimal configuration is two identical parallel branches, where one branch contains $10,\Omega$ in series with $40,\Omega$ and the other contains $20,\Omega$ in series with $30,\Omega$. The equivalent external resistance is

$$R_{\text{ext}} = 25,\Omega.$$

The resulting total heater power is

$$P_{\max} = \frac{(20,\text{V})^2 \cdot 25,\Omega}{(20,\Omega + 25,\Omega)^2} \approx 4.94,\text{W}.$$

$$\boxed{P_{\max} \approx 4.94,\text{W}}$$

Sanity Checks

The expression $P_{\text{ext}} = \mathcal{E}^2 R_{\text{ext}}/(r + R_{\text{ext}})^2$ has dimensions of $\text{V}^2/\Omega = \text{W}$, confirming consistency.

The maximum possible power transfer for a $20,\text{V}$ source with $20,\Omega$ internal resistance is $\mathcal{E}^2/(4r) = 400/(80) = 5,\text{W}$, so the obtained $4.94,\text{W}$ lies just below the theoretical upper bound, consistent with the imposed element constraints.

The most critical potential error would be assuming a $20,\Omega$ load is achievable without violating the $2,\text{W}$ limit, since that configuration would force higher branch currents and would drive at least one resistor above its thermal limit.