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.
-
0.113 A
- 0.160 A
- 150 A
- 1324 A
The correct answer is: A
Explanation
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