An inductor is connected to a sinusoidal voltage with an amplitude of 120v. A peak current 3.0A appears in the inductor.a) What is the maximum current if the frequency of the applied voltage is doubled?b) What is the inductive reactance at each of the two frequencies?I started it but can't calculate the frequency or know what formal to use to get max Currentthanks Jack
The impedance (or reactance) of an inductor is: Z jωL And this tells you the ratio of the voltage to current: Z V/I So if you re-arrange this for I, you get: I V / Z V / (jωL) Now you can see that if you double the frequency (which is the same as doubling ω) that the current will be halved. Part b should be straightforward using the equations above, you don't need to find the frequency (and actually, you can't in this case).
You do not need the actual frequencies to solve this problem. You can use the frequency ratios. Since the current is given as 3.0 Amps peak I will assume the 120 Volts is peak also. (b1) XL at original frequency .(707 x 120V) / (.707 x 3A) 40 Ohms XL at (2) x (original frequency) 2pi x 2f x L (b2) But 2pi and L are both constant therefore XL at 2f 80 Ohms (a) Maximum current when frequency is doubled V / XL 120V / 80 Ohm 1.5 Amps
