Question

An AC source producing emf
ε = ε₀ [cos (100 π s⁻¹)t + cos (500 π s⁻¹)t]
is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be i = i₁ cos [(100 π s⁻¹)t + φ₁) + i₂ cos [(500π s⁻¹)t + ϕ₂]. So,

(a) i₁ > i₂
(b) i₁ = i₂
(c) i₁ < i₂
(d) The information is insufficient to find the relation between i₁ and i₂.

Answer 1

(c) i₁ < i₂

The charge on the capacitor during steady state is given by,
$Q=C\epsilon ={\epsilon }_{0}C\left[\cos \left(100{\mathrm{\pi s}}^{-1}\right)t+\cos \left(500{\mathrm{\pi s}}^{-1}\right)t\right]$

The steady state current is, thus, given by,
$$\begin{gathered} i=\frac{dQ}{dt}={\epsilon }_{0}C\times 100\mathrm{\pi }\left[\sin \left(100{\mathrm{\pi s}}^{-1}\right)t\right]+{\epsilon }_{\mathit{0}}C\times 500\mathrm{\pi }\left[\sin \left(500{\mathrm{\pi s}}^{-1}\right)t\right] \\ \Rightarrow i=100C{\mathrm{\pi \epsilon }}_0\cos \left[\left(100{\mathrm{\pi s}}^{-1}\right)\mathrm{t}+{\varphi }_1\right]+500C{\mathrm{\pi \epsilon }}_0\cos \left[\left(500{\mathrm{\pi s}}^{-1}\right)+{\varphi }_2\right] \\ \Rightarrow i_1=100C{\mathrm{\pi \epsilon }}_0\&\mathit{}{i}_{\mathit{2}}=500C{\mathrm{\pi \epsilon }}_0 \\ \therefore i_2>i_1 \end{gathered}$$