Explanation:
Impedence, \(Z=\sqrt{R^{2}+X_{L}^{2}}\)
Current, \(I=\dfrac{V}{Z}=\dfrac{V}{\sqrt{R^{2}+X_{L}^{2}}}\)
Here, \(R = 10\,\Omega ,L = 2.0H,V = 120\;V\)
\(v = 60\;H\,z,\omega = 2\,\pi v = 120\,\pi \,rad{s^{ - 1}}\)
\({X_L} = \omega L = 240\,\pi \simeq 754\,\Omega \)
\(\therefore \quad I = \frac{{120}}{{\sqrt {{{10}^2} + {{754}^2}} }} = 0.16\;A\)