The circuit consists of a constant current source $I_0$ driving two parallel branches: one with an Inductor ($L$) and a switch ($S$), and the other with a Resistor ($R$).

• Before $t=0$: Switch $S$ is open. The entire current $I_0$ flows through the resistor $R$. Current in inductor $I_1(0) = 0$.
• After $t=0$: Switch $S$ is closed. The current $I_0$ divides between the inductor ($I_1$) and the resistor ($I_2$).

According to Kirchhoff’s Current Law (KCL) at the top node:

$$ I_1 + I_2 = I_0 \implies I_2 = I_0 – I_1 $$

Since the branches are in parallel, the potential difference across them is equal ($V_L = V_R$):

$$ L \frac{dI_1}{dt} = I_2 R $$

Substituting $I_2$ into the voltage equation:

$$ L \frac{dI_1}{dt} = (I_0 – I_1)R $$

Rearranging terms to separate variables:

$$ \frac{dI_1}{I_0 – I_1} = \frac{R}{L} dt $$

Integrating both sides with limits $t=0$ ($I_1=0$) to $t$:

$$ \int_{0}^{I_1} \frac{dI_1}{I_0 – I_1} = \int_{0}^{t} \frac{R}{L} dt $$

$$ -\ln(I_0 – I_1) \Big|_0^{I_1} = \frac{R}{L} t $$

$$ \ln\left(\frac{I_0 – I_1}{I_0}\right) = -\frac{R}{L} t \implies I_0 – I_1 = I_0 e^{-\frac{R}{L}t} $$

Using $I_2 = I_0 – I_1$, we get the current through the resistor:

$$ I_2(t) = I_0 e^{-\frac{R}{L}t} $$

The total heat dissipated ($H$) in the resistor is the integral of power ($I^2 R$) over time from $t=0$ to $\infty$:

$$ H = \int_{0}^{\infty} I_2^2 R \, dt $$

$$ H = \int_{0}^{\infty} \left( I_0 e^{-\frac{R}{L}t} \right)^2 R \, dt = I_0^2 R \int_{0}^{\infty} e^{-\frac{2R}{L}t} \, dt $$

Solving the integral:

$$ H = I_0^2 R \left[ \frac{e^{-\frac{2R}{L}t}}{-2R/L} \right]_{0}^{\infty} $$

$$ H = I_0^2 R \left( 0 – \frac{1}{-2R/L} \right) = I_0^2 R \left( \frac{L}{2R} \right) $$

$$ H = \frac{1}{2} L I_0^2 $$