Final answer:
The voltage across the capacitor at the instant when the inductor voltage is 80 V is 40 V because the capacitor and inductor voltages are 180° out of phase. The voltage across the resistor does not change instantaneously as it is in phase with the current and is thus 160 V.
Step-by-step explanation:
In an L-R-C series circuit, the instantaneous voltages across the inductor (VL), capacitor (Vc), and resistor (VR) are related due to Kirchhoff's voltage law, which states that the sum of the potential differences (voltage) around any closed network is zero. Given that the instantaneous voltage across the inductor is 80 V, and the voltage amplitudes are VL = 180 V, Vc = 120 V, and VR = 160 V, voltages across the capacitor and resistor can be calculated by making use of the fact that the voltages across the inductor and the capacitor are 180° out of phase.
Part A: At the instant when the voltage across the inductor VL is 80 V, the voltage across the capacitor Vc will be the amplitude Vc = 120 V minus the instantaneous voltage across the inductor since they are 180° out of phase. Therefore, the voltage across the capacitor at this instant is Vc = 40 V.
Part B: As the voltage across the resistor VR is in phase with the current, the instantaneous voltage across it remains the same as its amplitude, which is VR = 160 V.