An alternating source drives a series RLC circuit with an emf amplitude of 6.12

Question

An alternating source drives a series RLC circuit with an emf amplitude of 6.12 V, at a phase angle of +30.2°. When the potential difference across the capacitor reaches its maximum positive value of +5.08 V, what is the potential difference across the inductor (sign included)

An alternating source drives a series RLC circuit with an emf amplitude of 6.12 V, at a phase angle of +30.2°. When the potential difference across the capacitor reaches its maximum positive value of +5.08 V, what is the potential difference across the inductor (sign included)

Explanation / Answer

phase = 30.2 degrees

inductir voltage = VL = 5.08 V

emf = Vtot = 6.08 V

tan = (VL - VC)/Vot

tan 30.2 = (VL - 5.08)/6.08

VL = 8.619 V

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