Calculating Peak Current in Inductive Reactance of 0.1H @340V, 50Hz

[Roomie]

Homework Statement

What is the peak current when the voltage is applied across a pure inductance of value 0.1H?

We're told that the Peak Voltage is 340V, that the power supply is rated at 240V rms.
f=50hz

Homework Equations

X_L = jwL

w=2pi*f

The Attempt at a Solution

w=2pi*50 = 100pi

X_L = jwL = j*100pi*0.1 = 10pi*j

I've never done this before so I have no idea if this is correct or a stupid answer?

 

[cepheid]

It's correct so far, and it's not a stupid answer. However, you haven't finished solving the problem. The problem asks for the peak *current.*

In general, current equals voltage over impedance, right? I = V/Z.

For the case of an ideal inductor, the impedance consists entirely of reactance, right? There is no resistance. So:

Z = X_L

When you carry out the division V/X_L, you get the amplitude and phase of the sinusoidal current being driven across the inductor.

 

[Roomie]

It's correct so far, and it's not a stupid answer. However, you haven't finished solving the problem. The problem asks for the peak *current.*

In general, current equals voltage over impedance, right? I = V/Z.

For the case of an ideal inductor, the impedance consists entirely of reactance, right? There is no resistance. So:

Z = X_L

When you carry out the division V/X_L, you get the amplitude and phase of the sinusoidal current being driven across the inductor.

Thanks for your help, it's much appreciated!

So I get a final answer of 34/jpi ?
So it's an imaginary answer right? I'm relatively new to imaginary numbers too, so would this be the right way to write it? Thanks!

 

[cepheid]

Right, the full phasor current (which contains information about the amplitude and the phase of this AC current) is:

\tilde{I} = \frac{34}{j\pi} = -j\frac{34}{\pi}

I've just used the I with a tilde (squiggle) over it to represent the complex "phasor" current. However, the question is just asking you for the peak value of the current (in other words, its amplitude). So you don't need the -j part (which represents a phase shift of -\frac{\pi}{2} relative to the AC voltage). The trick to seeing why this is true is to rewrite this complex number in a form that makes it clearer what is its amplitude and what is its phase. Basically, -j can be rewritten as a complex exponential:

-j = e^{-j\frac{\pi}{2}}

If you don't understand why this is true, don't worry about it. You will learn it in due course. In any case, the current then becomes:

\tilde{I} = \frac{34}{\pi}e^{-j\frac{\pi}{2}}

So, the magnitude of this complex number (the factor in front of the exponential) represents the amplitude of the signal. The phase of the complex number (which is the argument of the exponential -- i.e. the thing multiplying the j in the exponent), represents the phase of the signal. At the end of the day, the complex number is just two real numbers: the amplitude and the phase, which is what makes it such a convenient and compact way to describe a sinusoid. Engineers like to write the above in "phasor notation" as:

\tilde{I} = \frac{34}{\pi} \angle -\frac{\pi}{2}

I hope this helps.