Kvant Physics Problem 315

A wire spiral is connected to an ideal constant-voltage mains source with voltage $V$.

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Problem

F315. A wire spiral connected to the mains electricity supply is heated by an electric current. Half of the spiral is cooled (for example, with water). How will this affect the amount of heat released by this half of the spiral? By the other half? By the entire spiral? Assume the supply voltage is constant.

Setup and Assumptions

A wire spiral is connected to an ideal constant-voltage mains source with voltage $V$. The spiral consists of two equal halves connected in series along the same conductor, so that the same current flows through both halves at any instant. Initially each half has the same electrical resistance $R$ at uniform temperature.

Half of the spiral is cooled, producing a change in its temperature and therefore in its resistance. The other half is not directly cooled but is affected only through the change in current and possible self-heating.

Let the cooled half be labeled 1 and the other half be labeled 2. After cooling, the resistance of the cooled half becomes $R_1 = R(1-\varepsilon)$, where $\varepsilon$ is a dimensionless parameter with $0<\varepsilon<1$. The resistance of the other half is taken as $R_2 = R$, assuming its temperature change due to redistribution is secondary in the leading-order analysis of electrical response.

The current is assumed to be steady, so thermal and electrical transients are neglected. The voltage source is ideal, so internal resistance of the supply is ignored. The temperature coefficient of resistance is not explicitly needed beyond its effect being encoded in $\varepsilon$.

The quantities of interest are the Joule heat powers $P_1$ and $P_2$ released in each half and the total power $P_{\text{tot}}$.

Physical Principles

Ohm’s law for a resistor of resistance $R_{\Sigma}$ under constant applied voltage $V$ gives the current

$$I = \frac{V}{R_{\Sigma}}.$$

For series-connected resistors, the total resistance is

$$R_{\Sigma} = R_1 + R_2.$$

Joule heating in a resistor is given by

$$P = I^2 R,$$

applied separately to each segment carrying the same current.

The total electrical power delivered by the source is

$$P_{\text{tot}} = VI = \frac{V^2}{R_{\Sigma}}.$$

Derivation

Before cooling, both halves have resistance $R$, so the total resistance is

$$R_{\Sigma 0} = 2R,$$

and the current is

$$I_0 = \frac{V}{2R}.$$

The initial power in each half is

$$P_0 = I_0^2 R = \left(\frac{V}{2R}\right)^2 R = \frac{V^2}{4R}.$$

After cooling, the resistances become $R_1 = R(1-\varepsilon)$ and $R_2 = R$. The total resistance becomes

$$R_{\Sigma} = R(1-\varepsilon) + R = R(2-\varepsilon).$$

The new current is

$$I = \frac{V}{R(2-\varepsilon)}.$$

The power released in the cooled half is

$$P_1 = I^2 R_1 = \frac{V^2}{R^2(2-\varepsilon)^2},R(1-\varepsilon) = \frac{V^2}{R},\frac{1-\varepsilon}{(2-\varepsilon)^2}.$$

The power in the other half is

$$P_2 = I^2 R_2 = \frac{V^2}{R^2(2-\varepsilon)^2},R = \frac{V^2}{R},\frac{1}{(2-\varepsilon)^2}.$$

The total power is

$$P_{\text{tot}} = P_1 + P_2 = \frac{V^2}{R},\frac{2-\varepsilon}{(2-\varepsilon)^2} = \frac{V^2}{R(2-\varepsilon)}.$$

Result

The initial power in each half is

$$P_0 = \frac{V^2}{4R}.$$

After cooling one half so that $R_1 = R(1-\varepsilon)$ while $R_2 = R$, the powers are

$$P_1 = \frac{V^2}{R},\frac{1-\varepsilon}{(2-\varepsilon)^2}, \quad P_2 = \frac{V^2}{R},\frac{1}{(2-\varepsilon)^2}, \quad P_{\text{tot}} = \frac{V^2}{R(2-\varepsilon)}.$$

Relative to the initial state, for small $\varepsilon$ one obtains

$$P_1 < P_0, \quad P_2 > P_0, \quad P_{\text{tot}} > \frac{V^2}{2R}.$$

Thus, the cooled half releases less heat than before, the other half releases more heat than before, and the total heat released by the spiral increases.

Sanity Checks

Dimensional consistency follows from $V^2/R$, which has units $\mathrm{W}$ since $\mathrm{V}^2/\Omega = \mathrm{W}$. Each expression for $P_1$, $P_2$, and $P_{\text{tot}}$ contains the same prefactor $V^2/R$ multiplied by a dimensionless function of $\varepsilon$, ensuring correct units.

In the limit $\varepsilon \to 0$, both halves become identical and the expressions reduce to

$$P_1 = P_2 = \frac{V^2}{4R}, \quad P_{\text{tot}} = \frac{V^2}{2R},$$

matching the initial uniform-temperature state.

For small positive $\varepsilon$, expansion of $P_2$ gives an increase proportional to the increase in current, since the factor $(2-\varepsilon)^{-2}$ grows faster than the resistance change alone can compensate. For $P_1$, the reduction in resistance competes with the increase in current, and the quadratic dependence on $I$ makes the net change negative at leading order in $\varepsilon$.

The most sensitive point is the coupling between temperature change and resistance distribution: assuming $R_2$ unchanged isolates the dominant electrical feedback from the cooled segment, while higher-order thermal feedback corrections affect only higher-order terms in $\varepsilon$ without changing the qualitative result.