Question

The following series L-C-R circuit, when driven by an emf source of angular frequency \(70\) kilo-radians per second, the circuit effectively behaves like

• A. purely resistive circuit
• B. series R-L circuit
• C. series R-C circuit
• D. series R-L-C circuit with \(R = 0\)

Hint:

If \(X_C > X_L\), circuit is capacitive and behaves like R-C. If \(X_L> X_C\), it behaves like R-L.

Answer 1

The Correct Option is C

Step 1: Read the given values.
From circuit diagram:
\[ L = 100\,\mu H = 100\times 10^{-6}H \]
\[ C = 1\,\mu F = 1\times 10^{-6}F \]
\[ R = 10\Omega \]
Given:
\[ \omega = 70\,k\,rad/s = 70\times 10^3\,rad/s \]
Step 2: Compute inductive reactance.
\[ X_L = \omega L = (70\times 10^3)(100\times 10^{-6}) \]
\[ X_L = 70\times 10^3 \times 10^{-4} = 7\Omega \]
Step 3: Compute capacitive reactance.
\[ X_C = \frac{1}{\omega C} = \frac{1}{(70\times 10^3)(1\times 10^{-6})} \]
\[ X_C = \frac{1}{70\times 10^{-3}} = \frac{1}{0.07} \approx 14.3\Omega \]
Step 4: Compare \(X_L\) and \(X_C\).
Since:
\[ X_C>X_L \]
Net reactance:
\[ X = X_L - X_C<0 \]
So circuit behaves capacitive.
Thus it behaves like a series R-C circuit.
Final Answer:
\[ \boxed{\text{series R-C circuit}} \]