An inductor L, capacitor C and resistor R are connected in series to an AC source. RMS voltages across L, C and R are respectively 40 V, 10 V and 40 V. If the AMPLITUDE (peak value) of current is 10√2 A, determine the total impedance.
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Question

An inductor L, capacitor C and resistor R are connected in series to an AC source. RMS voltages across L, C and R are respectively 40 V, 10 V and 40 V. If the AMPLITUDE (peak value) of current is 10√2 A, determine the total impedance.

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G1 Physics · Accepted Answer

Step 1: The given voltages across L, C and R are V_L=40 V, V_C=10 V and V_R=40 V. Step 2: The net reactive voltage is V_L−V_C=30 V, perpendicular to resistive voltage V_R in the phasor diagram. Step 3: Total RMS voltage is V = √[40²+30²] = 50 V. Step 4: Peak current is I₀=10√2 A, so RMS current is I=I₀/√2=10 A. Step 5: Impedance Z=V/I=50/10=5 Ω. Final answer: (D) 5 Ω.