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,

विकल्प

• i₁ > i₂

• i₁ = i₂

•  i₁ < i₂

• The information is insufficient to find the relation between i₁ and i₂.

Answer 1

i₁ < i₂

The charge on the capacitor during steady state is given by,
`Q = C epsilon = epsilon _0C [cos(100pis^-1)]`
the steady state current is, thus, given by,
`i =(dQ)/dt = epsilon _0Cxx100pi [sin(100pis^-1)] + epsilon_0Cxx 500pi [sin({00pis)]`
c`
⇒ i = 100Cepsilon cos  [(100pis^-1)] t +  Ø_1+500piepsilon_0 cos [(500pis^-1)] + Ø_2]`
⇒ `i_1 = 100Cpiepsilon_0 & i_2 = 500C_piepsilon_0`
∴ i_2 > i_1