The circuit shown below, will act as an ideal current source with respect to terminals A and B when frequency is :
4 rad/sec
What makes a source "ideal" as a current source. A practical source is a Norton equivalent — a current source with a shunt output impedance. It approaches the ideal current source when that output impedance becomes infinite, so that the current delivered is independent of whatever load is connected:
\(Z_{out}\to\infty\)
So the question is really: at what frequency does the impedance seen at A–B become infinite?
Step 1 — find the output impedance. Kill the independent source by shorting it. Looking back into A–B, the capacitor sits directly across the terminals, and the inductor also reaches those terminals through the short. The two are therefore in parallel:
\(Z_{out}=\dfrac{Z_LZ_C}{Z_L+Z_C}=\dfrac{(j\omega L)\left(\dfrac{1}{j\omega C}\right)}{j\omega L+\dfrac{1}{j\omega C}}=\dfrac{j\omega L}{1-\omega^{2}LC}\)
Step 2 — make it infinite. The denominator must vanish:
\(1-\omega^{2}LC=0 \quad\Longrightarrow\quad \omega_0=\dfrac{1}{\sqrt{LC}}\)
This is parallel resonance — the inductive and capacitive currents are equal and opposite, they circulate entirely within the LC loop, and none is drawn from outside. An ideal parallel LC at resonance is an open circuit.
Step 3 — substitute the values.
\(\omega_0=\dfrac{1}{\sqrt{\left(\tfrac{1}{16}\right)(1)}}=\dfrac{1}{\tfrac{1}{4}}=4\ \text{rad/s}\)
which is option 4.
Where the distractors come from. Option 3, 16 rad/s, is what you get by forgetting the square root — that is \(1/LC\), which has the units of ω2, not ω. Option 1 would need LC = 1. Option 2, ω = 0, is the DC case, where the inductor is a short and the capacitor an open — the output impedance is then zero, which is the condition for an ideal voltage source, the exact opposite of what is asked.
The dual worth remembering. Series resonance makes the impedance minimum (a short at the terminals) and so approximates an ideal voltage source; parallel resonance makes it maximum and so approximates an ideal current source. Both occur at the same \(1/\sqrt{LC}\) — it is the topology, not the frequency, that decides which behaviour appears.
Hence, the circuit acts as an ideal current source at 4 rad/sec.
Similar Questions
In a parallel RLC circuit shown in figure at resonance
A. Current I flows through the resistor
B. Total current flowing through LC combination is zero
C. Current flows in the inductor is zero
D. total current flowing through LC combination is 2 A
E. Current flows in the capacitor is zero
Choose the most appropriate answer from the options given below :
Important Questions from Electrical Resonance
Consider the following statements and choose the correct answer:
Statement 1: In an RLC series circuit, at half-power frequencies the magnitude of total reactance is equal to resistance.
Statement 2: In an RLC parallel circuit, at half-power frequencies the magnitude total susceptance is equal to conductance.
If the bandwidth of a band-pass filter increases:
What will be the bandwidth of a series resonant circuit provided it has an inductive reactance of 1000 Ohm, a capacitive reactance of 1000 Ohm and a resistance of 0.1 Ohm? It is also known that the resonant frequency is \(10^6\) rad/sec
The peak output of a tuned amplifier is at 6 MHz and has a quality factor of 60. The bandwidth and 3 dB frequencies shall be
A series RLC circuit has a bandwidth of 300 rad/sec at a resonant frequency of 3000 rad/sec when excited by a voltage source of 100 V. The inductance of the coil is 0.1 H. The value of R and the voltage across C are, respectively
