1. Steady State Heat Balance

In the steady state, the rate of electrical heat generation (power dissipated) in the resistor must equal the rate of heat transfer to the surroundings.

The power dissipated by the resistor is given by: $$P = \frac{V^2}{R}$$

The rate of heat transfer is given by: $$H = \beta(\theta – \theta_0)$$

Equating Power and Heat transfer ($P = H$): $$\frac{V^2}{R} = \beta(\theta – \theta_0)$$

2. Substituting Resistance Dependence

The resistance varies with temperature according to the relation $R = R_0 \{1 – \alpha(\theta – \theta_0)\}$. Let us define the temperature rise as $\Delta \theta = \theta – \theta_0$.

Substituting $R$ and $\Delta \theta$ into the energy balance equation: $$\frac{V^2}{R_0(1 – \alpha \Delta \theta)} = \beta \Delta \theta$$

Rearranging to form a quadratic equation in terms of $\Delta \theta$: $$V^2 = \beta R_0 \Delta \theta (1 – \alpha \Delta \theta)$$ $$V^2 = \beta R_0 \Delta \theta – \beta R_0 \alpha (\Delta \theta)^2$$ $$\beta R_0 \alpha (\Delta \theta)^2 – \beta R_0 \Delta \theta + V^2 = 0$$

Dividing the entire equation by $\beta R_0 \alpha$: $$(\Delta \theta)^2 – \frac{1}{\alpha}\Delta \theta + \frac{V^2}{\beta R_0 \alpha} = 0$$

3. Solving for Temperature ($\theta$)

Using the quadratic formula $\frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$ to solve for $\Delta \theta$: $$\Delta \theta = \frac{\frac{1}{\alpha} \pm \sqrt{\left(\frac{1}{\alpha}\right)^2 – 4(1)\left(\frac{V^2}{\beta R_0 \alpha}\right)}}{2}$$ $$\Delta \theta = \frac{1}{2\alpha} \left[ 1 \pm \sqrt{1 – \frac{4\alpha V^2}{\beta R_0}} \right]$$

To determine the correct sign, consider the limit $V \to 0$. In this case, the temperature rise $\Delta \theta$ should be zero.

• If we take the positive sign: $\Delta \theta \to \frac{1}{\alpha}$ (This implies $R \to 0$, which is physically incorrect).
• If we take the negative sign: $\Delta \theta \to 0$.

Let us define a reference voltage term $V_0$ to simplify the expression inside the square root. Let $V_0^2 = \frac{\beta R_0}{4\alpha}$, which gives $V_0 = \sqrt{\frac{\beta R_0}{4\alpha}}$. Substituting this back: $$\Delta \theta = \frac{1}{2\alpha} \left\{ 1 – \sqrt{1 – \left(\frac{V}{V_0}\right)^2} \right\}$$

Since $\theta = \theta_0 + \Delta \theta$, the steady state temperature is: $$\theta = \theta_0 + \frac{1}{2\alpha} \left\{ 1 – \sqrt{1 – \left(\frac{V}{V_0}\right)^2} \right\}$$

4. Solving for Current ($I$)

The current through the resistor is given by Ohm’s Law: $$I = \frac{V}{R} = \frac{V}{R_0(1 – \alpha \Delta \theta)}$$

From our expression for $\Delta \theta$: $$1 – \alpha \Delta \theta = 1 – \alpha \left[ \frac{1}{2\alpha} \left\{ 1 – \sqrt{1 – \left(\frac{V}{V_0}\right)^2} \right\} \right]$$ $$1 – \alpha \Delta \theta = 1 – \frac{1}{2} + \frac{1}{2}\sqrt{1 – \left(\frac{V}{V_0}\right)^2}$$ $$1 – \alpha \Delta \theta = \frac{1}{2} \left\{ 1 + \sqrt{1 – \left(\frac{V}{V_0}\right)^2} \right\}$$

Substituting this back into the current equation: $$I = \frac{V}{R_0 \cdot \frac{1}{2} \left\{ 1 + \sqrt{1 – \left(\frac{V}{V_0}\right)^2} \right\}}$$ $$I = \frac{2V}{R_0 \left\{ 1 + \sqrt{1 – \left(\frac{V}{V_0}\right)^2} \right\}}$$