Question 11 Solution

In a damped RLC circuit, at the instant when the current is maximum ($di/dt = 0$), the voltage across the inductor is zero ($V_L = L \frac{di}{dt} = 0$).
The loop equation is: $$ V_R + V_C + V_L = 0 \implies iR + \frac{q}{C} + 0 = 0 $$ Therefore, at maximum current $I_{max}$, the capacitor is not uncharged. Instead: $$ q = – I_{max} R C $$

The total electromagnetic energy $E$ stored in the system at this instant is the sum of the magnetic energy in the inductor and the potential energy in the capacitor: $$ E = \frac{1}{2} L I_{max}^2 + \frac{1}{2} \frac{q^2}{C} $$ Substituting $q = -I_{max} RC$: $$ E = \frac{1}{2} L I_{max}^2 + \frac{1}{2 C} (-I_{max} RC)^2 $$ $$ E = \frac{1}{2} L I_{max}^2 + \frac{1}{2} C R^2 I_{max}^2 $$ $$ E = \frac{1}{2} I_{max}^2 (L + C R^2) $$

The heat dissipated is the loss in total energy as the amplitude drops.
Initial Energy ($E_i$): With amplitude $I$: $$ E_i = \frac{1}{2} I^2 (L + C R^2) $$ Final Energy ($E_f$): With amplitude $0.5I$: $$ E_f = \frac{1}{2} (0.5I)^2 (L + C R^2) = 0.25 \left[ \frac{1}{2} I^2 (L + C R^2) \right] = 0.25 E_i $$

Heat Dissipated ($\Delta H$): $$ \Delta H = E_i – E_f = E_i – 0.25 E_i = 0.75 E_i $$ $$ \Delta H = 0.75 \times \frac{1}{2} I^2 (L + C R^2) $$ $$ \Delta H = 0.375 I^2 (L + C R^2) $$