1. **Problem Statement:** Given a series circuit with a voltage source $v_s = 20 \sin(10t + 30^\circ)$ V, a resistor $R = 4\ \Omega$, and an inductor $L = 0.2$ H, find the voltage across the inductor $v(t)$ and the current $i(t)$. 2. **Relevant Formulas and Concepts:** - The total voltage in the series circuit is the sum of voltage across the resistor and inductor: $$v_s = v_R + v_L$$ - Voltage across resistor: $$v_R = iR$$ - Voltage across inductor: $$v_L = L \frac{di}{dt}$$ - The circuit is driven by a sinusoidal source, so we use phasor analysis. 3. **Phasor Representation:** - Source voltage phasor: $$\tilde{V}_s = 20 \angle 30^\circ$$ - Angular frequency: $$\omega = 10\ \text{rad/s}$$ - Inductive reactance: $$X_L = \omega L = 10 \times 0.2 = 2\ \Omega$$ 4. **Impedance of the circuit:** $$Z = R + jX_L = 4 + j2$$ 5. **Calculate magnitude and phase of impedance:** $$|Z| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 4.472$$ $$\theta_Z = \tan^{-1}\left(\frac{2}{4}\right) = 26.565^\circ$$ 6. **Calculate current phasor:** $$\tilde{I} = \frac{\tilde{V}_s}{Z} = \frac{20 \angle 30^\circ}{4.472 \angle 26.565^\circ} = 4.472 \angle (30^\circ - 26.565^\circ) = 4.472 \angle 3.435^\circ$$ 7. **Convert current phasor back to time domain:** $$i(t) = 4.472 \sin(10t + 3.435^\circ)\ \text{A}$$ 8. **Calculate voltage across inductor phasor:** $$\tilde{V}_L = jX_L \tilde{I} = 2 \angle 90^\circ \times 4.472 \angle 3.435^\circ = 8.944 \angle (90^\circ + 3.435^\circ) = 8.944 \angle 93.435^\circ$$ 9. **Convert voltage across inductor back to time domain:** $$v(t) = 8.944 \sin(10t + 93.435^\circ)\ \text{V}$$ **Final answers:** - $$v(t) = 8.944 \sin(10t + 93.43^\circ)\ \text{V}$$ - $$i(t) = 4.472 \sin(10t + 3.43^\circ)\ \text{A}$$