Solution

Let the current through the load (and the resistor $R$) be $I=(12\mbox{ A})e^{i\omega t}$, where $\omega$ is the angular frequency of the source. Then, the voltage drops across the load and the resistor are $V_L=(120\mbox{ V})e^{i(\omega t-\pi/3)}$ and

$$V_R=IR=(120\mbox{ V})e^{i\omega t},$$

respectively. Consequently, the EMF of the source is $\mathcal{E}=V_L+V_R$, which becomes

$$\mathcal{E}=(120\mbox{ V})(e^{i\pi/6}+e^{-i\pi/6})e^{i(\omega t-\pi/6)}=(120\mbox{ V})\cdot2\cos\frac{\pi}{6}\cdot e^{i(\omega t-\pi/6).$$

Thus, the amplitude of the $EMF$ is $120\sqrt3\approx208$ V.