Assume that the voltage source produces the voltage of RMS value of 220 V and frequency 50 Hz. The reactances of the inductance and capacitance are
X L = 2 π f L , X C = 1 2 π f C . X_L=2\pi fL,\\
X_C=\frac{1}{2\pi fC}. X L = 2 π f L , X C = 2 π f C 1 . Since this is a parallel connection, apply Ohm's law to each element to find current:
I R = V R = 220 6 = 36.7 A . I L = V X L = 220 2 π ⋅ 50 ⋅ 7 = 0.1 A . I C = V X C = 220 ⋅ 42 2 π ⋅ 50 = 29.4 A . I_R=\frac{V}{R}=\frac{220}{6}=36.7\text{ A}.
\\
I_L=\frac{V}{X_L}=\frac{220}{2\pi\cdot50\cdot7}=0.1\text{ A}.
\\
I_C=\frac{V}{X_C}=\frac{220\cdot42}{2\pi\cdot 50}=29.4\text{ A}. I R = R V = 6 220 = 36.7 A . I L = X L V = 2 π ⋅ 50 ⋅ 7 220 = 0.1 A . I C = X C V = 2 π ⋅ 50 220 ⋅ 42 = 29.4 A . The source current is
I s = I R 2 + ( I L − I C ) 2 = 22 A . I_s=\sqrt{I_R^2+(I_L-I_C)^2}=22\text{ A}. I s = I R 2 + ( I L − I C ) 2 = 22 A .
Comments