Question

A capacitor, c=150mf, is connected to a source of alternating e.m.f of r.m.s value 300v and frequency 60hz. a, calculate the r.m.s value of the current. b, if this capacitor is replaced by an inductor of 60mh, calculate the r.m.s value of the resulting current

Answer 1

(a) Let's first find the capacitive reactance:

$$X_C=\dfrac{1}{\omega C}=\dfrac{1}{2\pi fC},$$ $$X_C=\dfrac{1}{2\pi\cdot60\ Hz\cdot150\cdot10^{-6}\ F}=17.7\ \Omega.$$

Finally, we can find the rms current:

$$I_{rms}=\dfrac{V_{rms}}{X_C}=\dfrac{300\ V}{17.7\ \Omega}=16.95\ A.$$

(b) Let's first find $X_L$:

$$X_L=\dfrac{1}{\omega L}=\dfrac{1}{2\pi fL},$$ $$X_L=\dfrac{1}{2\pi\cdot60\ Hz\cdot60\cdot10^{-6}\ H}=44.2\ \Omega.$$

Finally, we can find the rms current:

$$I_{rms}=\dfrac{V_{rms}}{X_L}=\dfrac{300\ V}{44.2\ \Omega}=6.78\ A.$$