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 = i 1 cos [(100 π s−1)t + φ1) + i 2 cos [(500π s−1)t + ϕ2]. So,

(a) i 1 > i 2
(b) i 1 = i 2
(c) i 1 < i 2
(d) The information is insufficient to find the relation between i 1 and i 2.

Answer 1

(c) i1 < i2

The charge on the capacitor during steady state is given by,

Q = C ε = ε0 C [cos (100πs-1)t + cos (500 πs-1)t]

The steady state current is, thus, given by,

i = dQ/dt

i = ε0 C [100π sin (100πs -1)t] + ε0 C [500π sin (500πs-1)t]

i = 100 C π ε0 cos [(100πs-1)t + ϕ1] + 500 C π ε0 cos [(500 πs-1) + ϕ2]

i1 = 100 C π ε0 & i2 = 500 C π ε0

∴i2>i1

Hope this helps...   : )

Answer 2

Explanation:

☢️PLEASE MARK AS BRAINLIEST☢️