A 2 μF capacitor is connected directly across a 150 V\(_{rms}\), 60 Hz A.C source. Calculate the r m s value of the current.
X\(_c\) = \(\frac{1}{2\pi FC}\)
X\(_c\) = \(\frac{1}{2 \times 3.142 \times 60 \times 2 \times 10^{-6}}\)
X\(_c\) = 1326.1192\(\Omega\)
I = \(\frac{\text{V}}{X_c}\)
I = \(\frac{150}{1326.1192}\) = 0.113A
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