An electric circuit connected to a wall socket (RMS voltage 120 V) is connected

Question

An electric circuit connected to a wall socket (RMS voltage 120 V) is connected to an inductor L = 205 mH and a capacitor C = 336 ?F, as well as two resistors R1 = 32.5 ? and R2 = 10.5 ?. (i) In the case where the alternating frequency is extremely low, so that the capacitor is very nearly fully charged and the inductor current settles to a steady value, what is the current passing through each resistor? (ii) In the case where the alternating frequency is extremely high, so that the capacitor stores little to no charge and the inductor generates a back-current equal to the forward current, what is the current passing through each resistor?

Explanation / Answer

1) if f 0 then XL 0 (short circuit) and XC = (open circuit)

hence, IR2 = 0 and IR1 = 120 / 32.5 = 3.69 A

2) when f is large => L is open and C is short , so :

IR1 = 0 and IR2 = 120 / 10.5 = 11.43 A

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