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 `V`. Find the times dependence of the conductors temperature t assuming the thermal power dissipated into surrounding space to vary as `q=k(T-T_0)` where `k` is a constant `T_0` is the surrounding temperature (equal to conductor's temperature at the initial moment).

Explanation

To find the time dependence of the conductor's temperature \( T(t) \) when a DC voltage \( V \) is applied, we can follow these steps:

### Step 1: Understand the Heat Generation and Loss
The power \( P \) generated in the conductor due to the applied voltage is given by:
\[ P = \frac{V^2}{R} \]
This power will increase the temperature of the conductor. However, the conductor also loses heat to the surroundings. The rate of heat loss \( Q \) is given by:
\[ Q = k(T - T_0) \]
where \( T_0 \) is the surrounding temperature.

### Step 2: Set Up the Heat Balance Equation
The change in heat energy \( dQ \) in the conductor can be expressed using its heat capacity \( C \):
\[ dQ = C \frac{dT}{dt} \]
The net heat change in the conductor can be described by the equation:
\[ \frac{V^2}{R} - k(T - T_0) = C \frac{dT}{dt} \]

### Step 3: Rearrange the Equation
Rearranging the equation gives:
\[ C \frac{dT}{dt} + k(T - T_0) = \frac{V^2}{R} \]
This can be rewritten as:
\[ \frac{dT}{dt} + \frac{k}{C}(T - T_0) = \frac{V^2}{RC} \]

### Step 4: Solve the Differential Equation
This is a first-order linear differential equation. To solve it, we can use an integrating factor. The integrating factor \( \mu(t) \) is given by:
\[ \mu(t) = e^{\int \frac{k}{C} dt} = e^{\frac{kt}{C}} \]
Multiplying through by the integrating factor:
\[ e^{\frac{kt}{C}} \frac{dT}{dt} + \frac{k}{C} e^{\frac{kt}{C}} (T - T_0) = \frac{V^2}{RC} e^{\frac{kt}{C}} \]

### Step 5: Integrate Both Sides
Integrating both sides with respect to \( t \):
\[ \frac{d}{dt}(T e^{\frac{kt}{C}}) = \frac{V^2}{RC} e^{\frac{kt}{C}} + \frac{kT_0}{C} e^{\frac{kt}{C}} \]
Integrating gives:
\[ T e^{\frac{kt}{C}} = \int \left( \frac{V^2}{RC} + \frac{kT_0}{C} \right) e^{\frac{kt}{C}} dt + C_1 \]

### Step 6: Solve for \( T \)
After integrating, we can express \( T \):
\[ T(t) = T_0 + \frac{V^2}{kR} (1 - e^{-\frac{kt}{C}}) \]

### Final Expression
Thus, the time dependence of the conductor's temperature is:
\[ T(t) = T_0 + \frac{V^2}{kR} (1 - e^{-\frac{kt}{C}}) \]