Solving for 'b': Current in a Loop w/ Bsin(1000t+b)

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Homework Statement


steady-state current through the loop will be of the form Bsin(1000t +b). What is the phase angle 'b'? in rads

Homework Equations

The Attempt at a Solution


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The answer is meant to be 0.46rads/s. However I am getting 9.65x10^-5??
I have no idea what I did wrong.

I first converted everything into the phasor domain. Then I used voltage division to get the voltage drop across the resistor. Then I used Ohms Law to get the current through the resistor which is the same as the current through the circuit.
 
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Check your calculation for the impedance of the inductor. The formula for the impedance of an inductor is not he same as that of a capacitor.

Don't be afraid to promote j's in the denominator to the numerator. If you have a value like ##1/0.25j##, then this can become ##-4j##.

You don't need to do any voltage division here. Simply sum up the impedances to find the total impedance. Then apply Ohm's law directly: ##I = E/Z## (where E is the applied potential difference across Z).
 
gneill said:
Check your calculation for the impedance of the inductor. The formula for the impedance of an inductor is not he same as that of a capacitor.

Don't be afraid to promote j's in the denominator to the numerator. If you have a value like ##1/0.25j##, then this can become ##-4j##.

You don't need to do any voltage division here. Simply sum up the impedances to find the total impedance. Then apply Ohm's law directly: ##I = E/Z## (where E is the applied potential difference across Z).
Ah.
Finally got it. Thank you soo much.
It says in my lecture notes that inductors can be described be V = jwLI or I = 1/jwL * V
It says nothing else. How do I know like what to use when? I mean say I am converting a 2H inductance into the phasor form, how do I know what to use?
 
##Z_L = j \omega L##

##Z_C = \frac{1}{j \omega C}##