A conductor has a temperature independent resistance R and a total heat capacity C. At the moment t = 0 it is connected to a DC voltage source of EMF V. Find the time dependence of the conductor's temperature T assuming the thermal power dissipated into surrounding space to vary as q = K(T − T0), where K is a constant, T0 is the surrounding temp, which is considered to be equal to conductor's initial temperature.

Key Ideas Needed

1. Joule Heating in a Resistor
   Electric power delivered to a resistor by a constant DC source is $P_{\text{elec}} = \frac{V^2}{R}$ because current $I = \frac{V}{R}$ and $P = VI$.

2. Heat Balance (First-Law for Lumped System)
   For a small object with heat capacity $C$, any net heat input changes its temperature: $C\,\frac{dT}{dt}= \text{(heat in)} - \text{(heat out)}$

3. Newton’s Law of Cooling for Heat Out
   The surrounding air is at $T_0$. If the object is at $T$, the rate of heat loss is $q_{\text{out}} = K\,(T - T_0)$ where $K$ (W/K) measures how easily heat escapes.

4. Linear First-Order Differential Equation
   Putting (1) and (3) into (2) gives a standard linear differential equation of the form $\frac{d\theta}{dt}+\frac{K}{C}\,\theta = \text{constant}$ where $\theta = T - T_0$. Such equations always give an exponential approach to the steady value.

5. Time Constant
   The denominator term $\tau = \frac{C}{K}$ tells how fast the system responds (large $C$ or small $K$ means slower changes).

Imagine a Toaster Wire

• Starting Point: At room temperature, the wire is cool just like your surroundings.
• Switch On: The moment you connect it to a battery, electric energy flows and the wire starts getting hot, exactly like the glowing coil in a toaster.
• Heat Story:
  1. Heat In: Electricity gives the wire a fixed amount of heat every second (because voltage $V$ and resistance $R$ stay the same).
  2. Heat Out: The hotter it gets, the faster it gives heat to the cool air around it. This is like blowing on hot soup— the hotter the soup, the faster it cools.
• Fight Between IN and OUT: At first, heat-in is bigger than heat-out, so the wire heats up. As it gets hotter, heat-out catches up. Finally they match and temperature stops rising.
• Math Picture: The temperature climbs quickly at the start, then slows down, and finally levels off. The curve looks like the way a balloon inflates fast and then less and less.

The job is to write that curve T(t) in a neat math formula.

Detailed Solution

1. Heat Balance Equation
   $C\,\frac{dT}{dt} = \frac{V^2}{R} - K\,(T - T_0)$

2. Define Temperature Rise
   $\theta = T - T_0 \quad\Rightarrow\quad \theta(0)=0$ $C\,\frac{d\theta}{dt} = \frac{V^2}{R} - K\,\theta$

3. Standard Linear ODE
   $\frac{d\theta}{dt} + \frac{K}{C}\,\theta = \frac{V^2}{R C}$

4. Integrating Factor Method
   Integrating factor $\mu(t)=\exp\Bigl(\frac{K}{C}t\Bigr)$. Multiply throughout and integrate: $\frac{d}{dt}\Bigl[\theta\,\mu(t)\Bigr] = \frac{V^2}{R C}\,\mu(t)$ $\theta\,e^{Kt/C} = \frac{V^2}{R C}\int_0^{t} e^{K t'/C}\,dt'$ $\theta\,e^{Kt/C} = \frac{V^2}{R K}\Bigl(e^{Kt/C}-1\Bigr)$

5. Isolate $\theta(t)$
   $\theta(t)=\frac{V^2}{K R}\Bigl(1 - e^{-Kt/C}\Bigr)$

6. Return to Absolute Temperature
   $\boxed{\;T(t)=T_0 + \frac{V^2}{K R}\Bigl(1 - e^{-t/\tau}\Bigr)\;},\qquad \tau = \frac{C}{K}$

7. Check Limits
   • $t \to 0$: $T \to T_0$ ✔️
   • $t \to \infty$: $T \to T_0 + \dfrac{V^2}{K R}$ ✔️