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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,
Options:
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Verified Answer
The correct answer is:
series \(R C\) circuit
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).
\(C=1 \mu \mathrm{F}, \quad R=10 \Omega\) 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).
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