Answer: I 2 = 0.4 A, I 1 = 0.02 A.
- A I<sub>2</sub> = 0.4 A, I<sub>1</sub> = 0.02 A
- B I<sub>2</sub> = 2 A, I<sub>1</sub> = 0.1 A
- C I<sub>2</sub> = 0.04 A, I<sub>1</sub> = 2 A
- D I<sub>2</sub> = 44 A, I<sub>1</sub> = 2.2 A
Correct answer: A. I<sub>2</sub> = 0.4 A, I<sub>1</sub> = 0.02 A
Explanation: V<sub>2</sub> = V<sub>1</sub>/20 = 2200/20 = 110 V. I<sub>2</sub> = P/V<sub>2</sub> = 44/110 = 0.4 A. By energy conservation: I<sub>1</sub> = I<sub>2</sub>/20 = 0.4/20 = 0.02 A.
In a pure inductor the current lags the voltage by 90° (energy is briefly stored in the magnetic field before current responds); in a pure capacitor the current leads the voltage by 90° (current must flow to build up charge before voltage rises) - the mnemonic "ELI the ICE man" keeps these straight.
Concept context
AC circuits, impedance, LCR series circuit, resonance, transformers, and power factor.