MCQ
An AC source producing emf $\in=\in_{0}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$ is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be $\text{i}=\text{i}_1\cos\Big[\big(100\pi\text{s}^{-1}\big)\text{t}+\phi_1\Big]+\text{i}_2\cos\Big[\big(500\pi\text{s}^{-1}\big)\text{t}+\phi_2\Big].$
  • A
    $i_1 > i_2$
  • B
    $i_1=i_2$
  • $i_1 < i_2$
  • D
    The information is insufficient to find the relation between $i_1$ and $i_2$

Answer

Correct option: C.
$i_1 < i_2$
$\text{Q}=\text{C}\in=\in_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$
$\text{i}=\frac{\text{dQ}}{\text{dt}}$
$\text{Q}=\text{C}\in=\in_{0}\text{C}\Big[\cos\big(100\pi\text{s}^{-1}\big)\text{t}+\cos\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$
$\in_0\text{C}\times100\pi\Big[\sin\big(100\pi\text{s}^{-1}\big)\text{t}\Big]\\+\in_0\text{C}\times500\pi\Big[\sin\big(500\pi\text{s}^{-1}\big)\text{t}\Big]$
$=100\text{C}\pi\in_0\cos\Big[\big(100\pi\text{s}^{-1}\big)\text{t}+\phi_1\Big]\\+500\text{C}\pi\in_0\cos\Big[\big(500\pi\text{s}^{-1}\big)\text{t}+\phi_2\Big]$
$\text{i}_1=100\pi\in_0\text{C}$ and $\text{i}_2=500\pi\in_0\text{C}$
$\text{i}_2>\text{i}_1$

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