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

Option (c)  is correct

Explanation:

The charge is given on the condenser during a steady state by,

$Q=C \varepsilon=\varepsilon_{0} C\left[\cos \left(100 \pi s^{-1}\right) t+\cos \left(500 \pi s^{-1}\right) t\right]$

Consequently, the constant state current is given by,

$\begin{aligned}&i=\frac{d Q}{d t}=\varepsilon_{0} C \times 100 \pi\left[\sin \left(100 \pi s^{-1}\right) t\right]+\varepsilon_{0} C \times 500 \pi\left[\sin \left(500 \pi s^{-1}\right) t\right]\\&i=100 C \pi \varepsilon_{0} \cos \left[\left(100 \pi s^{-1}\right) t+\phi_{1}\right]+500 C \pi \varepsilon_{0} \cos \left[\left(500 \pi s^{-1}\right)+\phi_{2}\right]\\&i_{1}=100 C \pi \varepsilon_{0} \& i 2=500 \mathrm{C} \pi \varepsilon_{0}\\&\therefore i_{2}>i_{1}\end{aligned}$