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One question from my college exam give to me this circuit, where the input is a sinusoid with a 1V amplitude. circuit

the question give to me a graph, frequency vs Amperies, that represent the current output value of resistor R1: output current R1

the first question is what's the nature of X? I think it's capacitive because the current increase with the frequency growing,in other words, the impedance of X decreases therefore X is a capacitor. But the second question get my mind blowing , it asks what's the values of R1 and R2.Could someone help me?? How do I find the R1 and R2 values from this AC circuit??

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    \$\begingroup\$ Well, assume that at very high frequencies the X value is 0 (or close enough to it.) What does that say about R1? \$\endgroup\$ – jonk May 9 at 19:13
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    \$\begingroup\$ So this looks like a high pass filter, right? Typically when you’re doing a frequency response, you use v/v and then convert to dBv to describe the voltage gain. But instead here it’s probably i/i and then you convert it to dBi to describe the current gain. I think the first thing you should do is find the transfer function which doesn’t require you to use any numbers. Your transfer function should look like something of a high pass filter. \$\endgroup\$ – user103380 May 9 at 19:23
  • \$\begingroup\$ @jonk, if X is a capacitor, the impedance tends to zero. So , for high frequencies, the impedance is so close of zero, so I can't conclude that in this situation the capacitor is a short circuit, Am I right?? \$\endgroup\$ – Lucas Vital May 9 at 21:30
  • \$\begingroup\$ @KingDuken very good suggestion. I solved the problem, it was only necessary analyse the transfer function in the frequency domain at two extremes, at high frequency and low frequencies \$\endgroup\$ – Lucas Vital May 9 at 22:02
  • \$\begingroup\$ @LucasVital X will be "open" at sufficiently low frequencies and therefore leaves you with a simple resistor voltage divider. X will be "closed" at sufficiently high frequencies and therefore shorts out R2, leaving only R1 powered by the circuit. So you have the current at two important regions informing you of the currents when it is R1 & R2 acting as resistor divider and again when it is only R1. That's a lot of information, looking at the chart. \$\endgroup\$ – jonk May 9 at 23:46
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Check your hypothesis (X is capacitance) at two extremes. At very low frequency, what is the impedance of a capacitor? At very high frequency, what is the impedance of a capacitor?

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Firstly it's necessary find the transfer function of the correspondent output. After that analyse the transfer function in the frequency domain at two extremes, at high frequency and low frequency and then you will easily find the R1 and R2 value.

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