Design a series resonant circuit with an input voltage of 12V< 0° to have the following specifications:A peak current 800 mA at resonanceA bandwidth of 180 HzA resonant frequency of 5200 HzFind the value of R, L, C and  fL and fH respectively.Group of answer choices15 ohm, 13 H, 0.71uF, 2600 Hz, 7800 Hz12 kohm, 13 H, 0.71pF, 2600 Hz, 5200 Hz12 kohm, 13 mH, 0.71pF, 2600 Hz, 7800 Hz15 ohm, 13 mH, 70.7nF, 5110 Hz, 5290 Hz

To design a series resonant circuit with the given specifications, we first need to find the values of resistance (R), inductance (L), and capacitance (C).

1. The peak current (I) at resonance is given as 800 mA or 0.8 A. The input voltage (V) is given as 12 V. We can use Ohm's law (V = IR) to find the resistance (R). Rearranging for R gives us R = V/I = 12V / 0.8A = 15 ohms.

2. The resonant frequency (fr) is given as 5200 Hz. The formula for resonant frequency is fr = 1 / (2π√(LC)). Rearranging for L gives us L = 1 / (4π²fr²C). But we don't have the value of C yet.

3. The bandwidth (BW) is given as 180 Hz. The formula for bandwidth in a series RLC circuit is BW = R / (2πL). We can rearrange this for L to get L = R / (2πBW). Substituting the known values gives us L = 15 ohms / (2π * 180 Hz) = 0.013 H or 13 mH.

4. Now we can find the value of C using the rearranged resonant frequency formula from step 2. Substituting the known values gives us C = 1 / (4π² * (5200 Hz)² * 13 mH) = 70.7 nF.

5. The lower and upper cut-off frequencies (fL and fH) can be found using the formulas fL = fr - BW/2 and fH = fr + BW/2. Substituting the known values gives us fL = 5200 Hz - 180 Hz/2 = 5110 Hz and fH = 5200 Hz + 180 Hz/2 = 5290 Hz.

So, the values of R, L, C, fL, and fH are 15 ohm, 13 mH, 70.7nF, 5110 Hz, and 5290 Hz respectively. This corresponds to the last option in your list of answer choices.