Question

I don't know how to do part iii) \( V_L \)
https://dl.dropbox.com/u/63664351/AC.PNG

Answer 1

Part a:
Calculate the inductive reactance of the coil:
\[X_L=2 \pi f L=2 \pi (50)(0.050H)=15.7 \Omega\]
Now calculate the total impedance of the circuit:\[Z=\sqrt{R^2+X_L^2}=\sqrt{5\Omega^2+15.7\Omega^2}=16.48\Omega=16.5 \Omega\]
Part b:
Calculate the load current using Ohm's law:\[I=\frac{V}{R}=\frac{240V}{16.5 \Omega}= 14.55A\]
Part c:
You can do this part using Ohm's law also...noting that the loads are in series:

For the resistor:
\[V=IR=(14.55A)(5\Omega)=72.75V\]
For the inductor:
\[V=IR=(14.55A)(15.7) \Omega)=228.4V\]

Answer 2

yeah but wheres the angle?

Answer 3

1)\[Z_L=R_L+i \omega L=R_L+i2 \pi f L=5+i15.71=\sqrt{5^2+15.71^{2}}/_{\arctan(15.71/5)}\]\[Z_L=16.48/_{72.34^o}\]
2)\[I_L=V_S/Z_L=[240/_{0^o}]/[16.48/_{72.34^o}]=14.56/_{-72.34^o}\]
3)\[V_R=RI_L=[5/_{0^o}][14.56/_{-72.34^o}]=72.8/_{-72.34^o} Volts\]
\[V_L=i \omega LI_L=[15.71/_{90^o}][14.56/_{-72.34^o}]=228.69/_{90^o-72.34^o}=228.69/_{17.66^o} Volts\]

Answer 4

Ahh...I forgot about the angles I guess but @CarlosGP did them for you :)