A series LRC circuit has L = 5 mH, C = 100 pF, and R = 1 kΩ. (Lab 5 uses the same values.)

(a) What is the resonant angular frequency ω0?

The resonance frequency for an LRC circuit is defined as

\(w_0=\frac{1}{\sqrt{LC}}\)

\(w_o=1.4\cdot10^6\) 

(b) What is the corresponding frequency f0 (in Hz)?

\(f=\frac{w_o}{2\pi f}\)

\(f=225KHz\)

(c) What is the magnitude of ZL, the impedance of the inductor, at this frequency?

\(Z_{inductor}=iwL\)

\(Z_{inductor}=i\cdot1.4\cdot10^6\cdot5\cdot10^{-3}=7000 \Omega\)

(d) What is the magnitude of ZC, the impedance of the capacitor, at this frequency? (Note that reactance XL ≥ 0 but reactance XC ≤ 0.)

\(Z_{capacitor}=\frac{1}{iwC}\)

\(Z_{capacitor}=\frac{1}{i\cdot1.4\cdot10^6\cdot100\cdot10^{-12}}=7142 \Omega\)

(e) If the input voltage to this circuit is a sinusoid and the output voltage is the voltage across the resistor, explain qualitatively why the amplitude of the output voltage is at a maximum value when the input voltage has frequency f = f0. (Hint: you can use a simple “voltage divider” argument.)

If we have an LRC circuit that is driven at the resonance frequency the inductor, the inductor and capacitor will create the lowest total impedance between them, causing the largest voltage drop across the resistor - maximizing V_out. 

Question 2

Match the four circuit diagrams with the four outputs. 

Circuit 2)  Only allows the negative part of the sign wave through the diode. That leaves us with only one choice- Graph B

Circuit 3) Here the current can flow both ways through the diode bridge. However, the output current will always flow in the same direction. So we will want the absolute value of the input function * some attenuation for the resistor. The only choice here is graph E. 

Circuit 1) - Circuit 1 and circuit 2 are the same just with the polarity of the diode switched. So if we know that that circuit 2 only lets the negative part through we should look for something of the same amplitude but the opposite polarity -  Graph C

Circuit 4) The only remaining choice is graph D. We can easily see that the direction of current flow is correct (lets the positive through but not the negative). So this choice makes sense as well.