In the given circuit, the angular frequency of the voltage source is \(70 \times 10^3 \mathrm{rad}…

In the given circuit, the angular frequency of the voltage source is \(70 \times 10^3 \mathrm{rad} \mathrm{s}^{-1}\). The circuit effectively behaves like,
  1. purely resistive circuit
  2. series RL circuit
  3. series \(R C\) circuit
  4. series \(L C\) circuit with \(R=0\)

Solution

Given, \(L=10 \mu \mathrm{H}\), \(C=1 \mu \mathrm{F}, \quad R=10 \Omega\) and angular frequency, \(\omega=70 \times 10^3 \mathrm{rad} \mathrm{s}^{-1}\) Now, the impedance of series LCR circuit, \(\begin{gathered} Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} \\ \Rightarrow \\ Z=\sqrt{10^2+\left[70 \times 10^{+3} \times 10 \times 10^{-6}-\frac{1}{70 \times 10^{+3} \times 1 \times 10^{-6}}\right]} \\ \Rightarrow \quad Z=\sqrt{100+\left[0.7-\frac{100}{7}\right]^2} \\ \Rightarrow \quad Z=\sqrt{100+(-1358)^2} \end{gathered}\) Here, negative sign shows that the circuit properties are more likely to be capacitive than the inductive i.e., \(\Rightarrow \quad X_C > X_L\) So, the circuit behaves like series \(R C\) circuit. Hence, the correct option is (c).

Asked in: AP EAMCET 2019 (23 Apr Shift 1)

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