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HNC electrical engineering

Im struggling with question D.
Can someone help please?
a) The resonant frequency of the circuit. b) The magnitude of the circuit current at resonance. c) The magnitude of the voltage across the inductor at resonance.
d) Account for the magnitude of the voltage across the inductor, given that the supply voltage is only 10V?
e) Sketch a phasor diagram showing the relative phases and magnitudes of the current and all the voltages in the above circuit.
f) A fault has occurred in the above circuit and the inductor needs to be replaced. However, the only inductor available has a value of 18 mH. The variable capacitance is adjusted to bring the circuit back to its original resonant frequency. What value of capacitance does the circuit now possess? when
V = 10volts
R = 10 ohms
L = 10mH
C = 25.33nF
Reply 1
Same here i think its asking why the answer to C is higher than the supply voltage, that's how I'm reading it
Reply 2
Did you have any luck answering this question?
Reply 3
probably a bit late to the party but i read question D the same, asking why C is so high.

took me a while to figure out how to answer as it was a bit confusing but this is basically what im going with:

the voltage across the inductor is 628.32v because when taken alone the resistance is XL= 2pifl so XL x I = 628.32 volts
when taken as a circuit the inductor and capacitor are equal and opposite Xc =1/2pifc = 628.32 ohms, They cancel each other out making the circuit resistance 10 ohms only which makes the voltage 1A.

will obviously try to make it a bit more detailed about why they cancel each other out

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