RC and RL(frequency and potential difference)

AI Thread Summary
As frequency increases, the potential difference across a capacitor decreases due to its reactance, which is inversely proportional to frequency. Conversely, for an inductor, the potential difference increases with frequency because its reactance is directly proportional to frequency. The time constant and period can influence the potential difference across both components by affecting how quickly they respond to changes in current. Understanding impedance as a complex number that combines resistance and reactance is crucial for analyzing AC circuits. Mastery of these concepts is essential for solving related problems in electrical engineering.
SAT2400
Messages
68
Reaction score
0

Homework Statement


1.(RC in AC)What will happen to the potential difference on the capacitor as the frequency increases? In terms of the reactance of the capacitor, why might this happen?

2.(RL in AC) What will happen to the potential difference on the inductor as the frequency increases? In terms of the reactance of the inductor, why might this happen?

3. How might time constant and period determine the inductors potential difference?

4. How might time constant and period determine the capacitors potential difference?

Homework Equations


XL= 2pif
V=IR
I= V/Z


The Attempt at a Solution


I do not know how to find the relationships between f and potential difference on both capacitor and inductor..What equations should I use?? Please help me answering these 4 questions! Thanks
 
Physics news on Phys.org
Hi SAT2400

For part 1, you know the impedance of a capacitor, right? You also know the impedance of a resistor (which is just its resistance). You know that they are in series (I assume?). If so, then you know that at any instant, the voltage drop across each has to add up to the voltage of the source (like in any series circuit). Does that help?
 
hmmm,,not quite...

Can you explain more in detail!?T_T

and how do I know the impedence? is it the resistance??

I have got values for Xc(ohm), Z(ohm), I(Amp), Vc(V), Vr(V) and Vs.
THank you
 
Okay, well the impedance, Z, of a component is just a complex number that describes the relationship between the complex (or "phasor") voltage across that component and the complex current through it. In general, this complex impedance has a real part, R, and an imaginary part, X, such that:

Z = R + jX

The real part, R, is just the resistance of the comonent, and the imaginary part, X, is called the "reactance." In the case of a capacitor or an inductor, there is NO resistance (no real part). The impedance is purely imaginary (or purely "reactive", to use the EE jargon):

ZC = jXC.

The nice thing about impedances, is that they allow us to use the same circuit analysis techniques for AC circuits as we do for DC circuits. If two components are in series, the total impedance is just the sum of the individual impedances: Z = Z1 + Z2 etc. The fact that the voltages and currents are oscillatory functions of time (which could potentially make things complicated) is neatly handled by the complex numbers and the phase information they contain. If you are learning about this stuff, all of what I just said should be in your notes or your book somewhere, and you should have an expression for XC as a *function* of omega so that you know how the impedance of a capacitor changes with frequency.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top