Let's solve the problem step by step: --- ## **a) Write an expression for \( v_s(t) \), the source voltage.** First, let's analyze the provided waveform: ### **Step 1: Amplitude** - The peak value is \( V_m = 18 \) V. ### **Step 2: Period and Frequency** - From the plot, the period \( T \) is the time between two consecutive peaks. - The first peak is at \( t = 15 \) ms, the next is at \( t = 55 \) ms. - So, \( T = 55\,\text{ms} - 15\,\text{ms} = 40\,\text{ms} = .04\,\text{s} \). \[ f = \frac{1}{T} = \frac{1}{.04} = 25\,\text{Hz} \] \[ \omega = 2\pi f = 2\pi \times 25 = 50\pi\,\text{rad/s} \] ### **Step 3: Phase** - At \( t = \), \( v_s() = \) and it's rising, which matches a sine function with zero phase shift. ### **Step 4: Expression** \[ \boxed{ v_s(t) = 18 \sin(50\pi t) } \] where \( t \) is in seconds. --- ## **b) Find the current flowing into the load** ### **Step 1: Impedance of the Load** The load is a parallel RC circuit: - \( R = 100\, \Omega \) - \( C = 25\, \mu\text{F} = 25 \times 10^{-6}\, \text{F} \) The total admittance \( Y \) is: \[ Y = \frac{1}{R} + j\omega C \] \[ Y = \frac{1}{100} + j(50\pi)(25 \times 10^{-6}) \] \[ Y = .01 + j(.00393) \] Impedance \( Z \) is: \[ Z = \frac{1}{Y} \] But we can express the current in the load as: \[ i(t) = \frac{v_s(t)}{Z} \] ### **Step 2: Phasor Analysis** Let \( v_s(t) = 18\sin(50\pi t) \). We convert this to phasor form: \( V_s = 18 \angle ^\circ \). Admittance in magnitude and phase: \[ |Y| = \sqrt{(.01)^2 + (.00393)^2} = .01075 \] \[ \theta_Y = \arctan\left(\frac{.00393}{.01}\right) = .375~\text{rad} \] So, \[ I = V_s \times Y = 18 \angle ^\circ \times .01075 \angle .375 \] \[ |I| = .1935~\text{A} \] \[ \text{Phase} = .375~\text{rad} \] So, in time domain: \[ \boxed{ i(t) = .194 \sin(50\pi t + .375) } \] --- ## **Final Answers** ### **Part (a):** \[ \boxed{ v_s(t) = 18\sin(50\pi t) } \] ### **Part (b):** \[ \boxed{ i(t) = .194\sin(50\pi t + .375) } \] where \( t \) is in seconds and the phase is in radians (between \(-\pi\) and \(\pi\)).