Question

The diode in the circuit is ideal. The current source $i_s(t)=\pi\sin(3000\pi t)$ mA. The magnitude of the average current flowing through the resistor $R$ is _____ mA (rounded off to two decimal places).

Hint:

For half-wave rectification, the average current is given by $I_{avg} = I_m/\pi$. Always check if the diode is ideal, as it decides conduction.

Answer 1

Correct Answer: 0.95

With an ideal diode, current flows through $R$ only during the positive half–cycles: \[ i_R(t)= \begin{cases} I_m\sin(\omega t), & 0<t<\tfrac{T}{2} \\ 0, & \tfrac{T}{2}<t<T \end{cases} \] where $I_m=\pi$ mA and $\omega=3000\pi$ rad/s. Average current: \[ I_{\text{avg}}=\frac{1}{T}\int_0^{T/2} I_m\sin(\omega t)\,dt =\frac{I_m}{T\omega}\Big[1-\cos(\omega T/2)\Big] =\frac{I_m}{2\pi} . 2 =\frac{I_m}{\pi}=1.00\ \text{mA}. \]