A coil with negligible resistance has 62.8 V across it with 0.01 A of current. How much is X_L?
Question 19.4: A coil with negligible resistance has 62.8 V across it with ...
X_L=\frac{V_L}{I_L}=\frac{62.8V}{0.01 A}
=6280 \Omega
Related Answered Questions
L=\frac{X_L}{2\pi f}=\frac{1}{2\pi f}\times...
L=\frac{X_L}{2\pi f}=\frac{6280}{6.28\times...
f=\frac{1}{2\pi L}\times X_L=\frac{1}{6.28\...
a. For a 10-H L,
X_L=2\pi fL=6.28\times 60\...
a. At 1 MHz,
X_L=2\pi fL=6.28\times 1\times...
X_L=2 \pi fL
=6.28\times 41...