Question

An alternating current source is connected in the following figure (a series combination of an inductor \(L\), a capacitor \(C\) and a resistor \(R\) across an AC source of rms voltage \(V\)). Determine the value of current flowing in the resistance \(R\) in the condition of resonance.

Hint:

At resonance \(X_L=X_C\), so the impedance drops to just \(R\). Apply \(I=V/Z\).

Answer 1

Step 1: Recall what happens at series resonance.
In a series \(L\)-\(C\)-\(R\) circuit driven at its resonant (natural) frequency, the voltage across the inductor and the voltage across the capacitor are equal in magnitude but opposite in phase, so together they cancel out completely.

Step 2: Reduce the effective circuit.
Because the inductor and capacitor voltages cancel, the whole source voltage \(V\) appears across the resistor alone. The net phase angle between voltage and current becomes zero, meaning the circuit is purely resistive at this frequency.

Step 3: Express the resonant angular frequency (background).
The cancellation \(X_{L}=X_{C}\) happens when \(\omega L=\dfrac{1}{\omega C}\), i.e. at \(\omega_{0}=\dfrac{1}{\sqrt{LC}}\). At this single frequency the reactive part vanishes.

Step 4: Compute the current directly.
With only \(R\) opposing the current, Ohm's law for the rms values gives
\[ I=\frac{V}{R} \]
This is the largest possible current for the given source, which is why resonance is also called the condition of maximum current.

Result:
\[\boxed{I=\dfrac{V}{R}}\]