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QPage 329: Q2: HC Verma: Concepts of Physics - Volume 2 Alternating Current Page 329: Q2 Solution: An AC source producing emf is connected in series with a…

An AC source producing emf ε=ε0[cos⁡100π s−1)t+cos⁡500π s−1)t]\varepsilon = \varepsilon_0 [\cos100 \pi \text{ s}^{-1})t + \cos 500 \pi \text{ 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=i1cos⁡[100π s−1)t+ϕ1]+i2cos⁡[500π s−1)t+ϕ2]i = i_1 \cos[100 \pi \text{ s}^{-1})t + \phi_1] + i_2 \cos[ 500 \pi \text{ s}^{-1})t + \phi_2]. So,

Multiple Choice Options

  • A i1>i2i_1 > i_2
  • B i1=i2i_1 = i_2
  • C i1<i2i_1 < i_2
  • D The information is insufficient to find the relation between i1i_1 and i2i_2.

Step-by-Step Verified Solution

**Answer:** (a) i_ > i_. **Step-by-Step Explanation:** The impedance is lower for lower frequencies in an RC circuit. So the current amplitude at 100� is larger than at 500�.

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