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An AC source with emf V = V0*(sin(wt) + sin(3wt)) is connected in series with a capacitor C and a resistor R. The resulting current is expressed as i = i1*sin(wt + phi1) + i2*sin(3wt + phi2). Which of the following correctly compares i1 and i2?
- i1 > i2
- i1 = i2
- i1 < i2
- any of the above may be true
Correct answer: i1 < i2
Solution
At the higher frequency 3w, the capacitive reactance X_C = 1/(3wC) is smaller than at w, giving a lower total impedance Z2 < Z1 and hence a larger current amplitude i2 > i1.
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