An AC source generating an emf E = E ₀ [ ! ((100 s ⁻¹)t ) + ! ((500 s ⁻¹)t ) ] is connected in series with a resistor and a capacitor. The steady-state current flowing in the circuit is observed to be i = i₁ ! [(100 s ⁻¹)t + ₁ ] + i₂ ! [(500 s ⁻¹)t + ₂ ] .

Step-by-step solution

The given emf has two frequency components: ₁ = 100 s ⁻¹ and ₂ = 500 s ⁻¹ . The impedance of an RC series circuit at an angular frequency is given by Z = R^2 + X_C^2 = R^2 + 1 ^2 C^2 . The amplitude of the steady-state current at frequency is i₀ = E ₀ Z = E ₀ R^2 + 1 ^2 C^2 . Since ₁ 1 ₂^2 C^2 . This implies that the impedance at ₁ is greater than the impedance at ₂ , i.e., Z₁ > Z₂ . Therefore, the current amplitude at ₁ is less than the current amplitude at ₂ , giving i₁ < i₂ .