Find the Inductance

From the expressions given, we see that the voltage leads the current by an angle of $\varphi = 0 - (-\dfrac{\pi}{3}) = \dfrac{\pi}{3}$ . Angle $\varphi$ is also the argument of the complex impedance formed by the series RL netrwork. Maximum values of $v(t)$ and $i(t)$ are $V_m = 10 \si{V}$ and $I_m = 3 \si{A}$ , so the RL impedance magnitude is $Z = \dfrac{V_m}{I_m} = \dfrac{10}{3} \si{\Omega}$ . The reactance of the inductor can be calculated as $X_L = Z \sin{\varphi} = \dfrac{10}{3} * \dfrac{\sqrt{3}}{2} \si{\Omega} = \dfrac{5\sqrt{3}}{3} \si{\Omega}$ , and from that we can determine the value of inductance: $L = \dfrac{X_L}{\omega} = \dfrac{5\sqrt{3}}{360\pi} \si{H} \approx 7.657 \si{mH}$

where $\omega = 120\pi \si{rad/s}$ is the angular frequency of the voltage, and it is derived from the given expression of $v(t)$ .

The impedance of the circuit is given by:

$\begin{aligned} Z & = \frac {v(t)}{i(t)} \\ \implies R + j {\color{#3D99F6} \omega} L & = \frac {10 \angle 0}{3 \angle \left(- \frac \pi 3\right)} & \small \color{#3D99F6} \text{where }\omega = 120 \pi \\ R + j {\color{#3D99F6} 120\pi} L & = \frac {10}{3 \left(\frac 12 - j \frac {\sqrt 3}2\right)} & \small \color{#3D99F6} \text{Multiply by }\frac {\frac 12 + j \frac {\sqrt 3}2} {\frac 12 + j\frac {\sqrt 3}2} \\ & = \frac {10\left(\frac 12 + j\frac {\sqrt 3}2\right)}3 \\ & = \frac 53 \left(1 + j \sqrt 3 \right) & \small \color{#3D99F6} \text{Equating the imaginary parts} \\ \implies 120 \pi L & = \frac 5{\sqrt 3} \\ L & = \frac 5{\sqrt 3 \times 120 \pi} \\ & \approx \boxed{7.657} \text{ mH} \end{aligned}$

Another solution exploits Kirchhoff's Voltage Law for this series RL circuit:

$v(t) = Ri(t) + Li'(t);$

or $10sin(120\pi t) = R[3sin(120\pi t - \frac{\pi}{3})] + L[360\pi cos(120\pi t - \frac{\pi}{3})];$

or $10sin(120\pi t) = 3R[sin(120\pi t)cos(\frac{\pi}{3}) - cos(120\pi t)sin(\frac{\pi}{3})] + 360\pi L[cos(120\pi t)cos(\frac{\pi}{3}) + sin(120\pi t)sin(\frac{\pi}{3})];$

or $10sin(120\pi t) = 3R[sin(120\pi t)(\frac{1}{2}) - cos(120\pi t)(\frac{\sqrt{3}}{2})] + 360\pi L[cos(120\pi t)(\frac{1}{2}) + sin(120\pi t)(\frac{\sqrt{3}}{2})].$

By matching coefficients for the sine and the cosine terms, we now obtain the following system of linear equations:

$10 = \frac{3}{2}R + 180\pi \sqrt{3}L;$ (i)

$0 = -\frac{3\sqrt{3}}{2}R + 180\pi L;$ (ii)

which solving $R$ for $L$ in (ii) gives $R = \frac{120\pi}{\sqrt{3}} \cdot L$ . Substituting this value into (i) ultimately yields $\boxed{L = \frac{1}{24\pi \sqrt{3}} = 7.657 \times 10^{-3}}$ Henries.