Help - PH5 Question
So I'm stuck on a question:
i) By considering reactances, explain why doubling R has little effect on the current at high frequencies.
I know that the reactances are:
Inductor= 2 x pi x f x L
Capacitor= 1 / 2 x pi x f x C
But I can't figure out why the resistance doesn't have an effect.
Someone help please, thanks
i) By considering reactances, explain why doubling R has little effect on the current at high frequencies.
I know that the reactances are:
Inductor= 2 x pi x f x L
Capacitor= 1 / 2 x pi x f x C
But I can't figure out why the resistance doesn't have an effect.
Someone help please, thanks
Original post by ScottGBR855
So I'm stuck on a question:
i) By considering reactances, explain why doubling R has little effect on the current at high frequencies.
I know that the reactances are:
Inductor= 2 x pi x f x L
Capacitor= 1 / 2 x pi x f x C
But I can't figure out why the resistance doesn't have an effect.
Someone help please, thanks
i) By considering reactances, explain why doubling R has little effect on the current at high frequencies.
I know that the reactances are:
Inductor= 2 x pi x f x L
Capacitor= 1 / 2 x pi x f x C
But I can't figure out why the resistance doesn't have an effect.
Someone help please, thanks
The question doesn't say no effect. It says "little" effect.
Does this relate to a particular circuit with a particular resistance and inductance in it?
Inductive reactance increases with frequency (your formula says this) so as you increase frequency this gets larger while the pure resistance of a resistor stays the same.
So as you increase frequency, any change in the value of R will have less effect so long as the value of R is small compared with the inductive reactance. Without a circuit or actual values it's difficult to be more specific.
As for capacitative reactance, that gets smaller with increasing frequency, so it is not the case for capacitance.
This is why I asked for more detail.
(edited 10 years ago)
Thanks for the reply.
It's a RLC circuit, with a resonance graph.
It gives the capacitor as 1.6x10^-6 F, inductor as 0.4H , 6v rms across the circuit, impedance is 800 ohms.
200hz is the resonance frequency (highest peak).
It's a 2 mark question, but it doesn't want any calculations just an explanation.
What I can't understand is why doubling the resistance wouldn't have a big effect?
Impedance = ( ( XL - XC )^2 + R^2 )^-1/2
If R was doubled the impedance would increase, and .. Impedance = Vrms / Irms .. therefore Irms = Vrms / Impedance, meaning that it would be smaller so has little effect but is it little at high frequencies because the reactances reduce each other?
It's a RLC circuit, with a resonance graph.
It gives the capacitor as 1.6x10^-6 F, inductor as 0.4H , 6v rms across the circuit, impedance is 800 ohms.
200hz is the resonance frequency (highest peak).
It's a 2 mark question, but it doesn't want any calculations just an explanation.
What I can't understand is why doubling the resistance wouldn't have a big effect?
Impedance = ( ( XL - XC )^2 + R^2 )^-1/2
If R was doubled the impedance would increase, and .. Impedance = Vrms / Irms .. therefore Irms = Vrms / Impedance, meaning that it would be smaller so has little effect but is it little at high frequencies because the reactances reduce each other?
Original post by ScottGBR855
Thanks for the reply.
It's a RLC circuit, with a resonance graph.
It gives the capacitor as 1.6x10^-6 F, inductor as 0.4H , 6v rms across the circuit, impedance is 800 ohms.
200hz is the resonance frequency (highest peak).
It's a 2 mark question, but it doesn't want any calculations just an explanation.
What I can't understand is why doubling the resistance wouldn't have a big effect?
Impedance = ( ( XL - XC )^2 + R^2 )^-1/2
If R was doubled the impedance would increase, and .. Impedance = Vrms / Irms .. therefore Irms = Vrms / Impedance, meaning that it would be smaller so has little effect but is it little at high frequencies because the reactances reduce each other?
It's a RLC circuit, with a resonance graph.
It gives the capacitor as 1.6x10^-6 F, inductor as 0.4H , 6v rms across the circuit, impedance is 800 ohms.
200hz is the resonance frequency (highest peak).
It's a 2 mark question, but it doesn't want any calculations just an explanation.
What I can't understand is why doubling the resistance wouldn't have a big effect?
Impedance = ( ( XL - XC )^2 + R^2 )^-1/2
If R was doubled the impedance would increase, and .. Impedance = Vrms / Irms .. therefore Irms = Vrms / Impedance, meaning that it would be smaller so has little effect but is it little at high frequencies because the reactances reduce each other?
At a frequency of 200Hz, if that is the resonant frequency of the circuit, the impedance is a minimum. Then you just have the pure resistance of the resistor in your V=IR.
This gives the maximum current. The reactances XC and XL "cancel out".
At very high (and very low) frequencies the current is very small because the total impedance of the circuit is very large, due to the reactance of L at high frequency and C at low frequency.
So the simple, non maths, answer is that at high frequencies the resistance of R (which doesn't change) is small compared with the overall impedance of the circuit. So doubling R adds little to the overall impedance of the circuit and means there is little change in the current.
Original post by Stonebridge
At a frequency of 200Hz, if that is the resonant frequency of the circuit, the impedance is a minimum. Then you just have the pure resistance of the resistor in your V=IR.
This gives the maximum current. The reactances XC and XL "cancel out".
At very high (and very low) frequencies the current is very small because the total impedance of the circuit is very large, due to the reactance of L at high frequency and C at low frequency.
So the simple, non maths, answer is that at high frequencies the resistance of R (which doesn't change) is small compared with the overall impedance of the circuit. So doubling R adds little to the overall impedance of the circuit and means there is little change in the current.
This gives the maximum current. The reactances XC and XL "cancel out".
At very high (and very low) frequencies the current is very small because the total impedance of the circuit is very large, due to the reactance of L at high frequency and C at low frequency.
So the simple, non maths, answer is that at high frequencies the resistance of R (which doesn't change) is small compared with the overall impedance of the circuit. So doubling R adds little to the overall impedance of the circuit and means there is little change in the current.
Ok thanks for your help. I asked my teacher today and he just wrote out the equation:
He said if (XL - XC) was 1000 and R was 1. Then if R was doubled to 2 it still wouldn't have a big effect as it's so small compared to the reactances (1000) . The difference between 1002 and 1001 is tiny.
Yes if R was 1 ohm. It depends on R, of course.
A value of R was not provided in your posts, but the whole point here is the relative size of R compared with the impedances of the other components.
You asked for a non calculation answer so you have to answer in general terms. Assuming a value of R is not really in the spirit of "non calculation".
Can I also say that it is much easier for us to answer questions when they are posted in their entirety, rather than just part questions or the poster's paraphrase of the question.
Anyway, you have your answer now. Yes?
A value of R was not provided in your posts, but the whole point here is the relative size of R compared with the impedances of the other components.
You asked for a non calculation answer so you have to answer in general terms. Assuming a value of R is not really in the spirit of "non calculation".
Can I also say that it is much easier for us to answer questions when they are posted in their entirety, rather than just part questions or the poster's paraphrase of the question.
Anyway, you have your answer now. Yes?
(edited 10 years ago)
Original post by Stonebridge
Yes if R was 1 ohm. It depends on R, of course.
A value of R was not provided in your posts, but the whole point here is the relative size of R compared with the impedances of the other components.
You asked for a non calculation answer so you have to answer in general terms. Assuming a value of R is not really in the spirit of "non calculation".
Can I also say that it is much easier for us to answer questions when they are posted in their entirety, rather than just part questions or the poster's paraphrase of the question.
Anyway, you have your answer now. Yes?
A value of R was not provided in your posts, but the whole point here is the relative size of R compared with the impedances of the other components.
You asked for a non calculation answer so you have to answer in general terms. Assuming a value of R is not really in the spirit of "non calculation".
Can I also say that it is much easier for us to answer questions when they are posted in their entirety, rather than just part questions or the poster's paraphrase of the question.
Anyway, you have your answer now. Yes?
Haha yeah, thanks mate.
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