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: Identify the circuit and the resonance condition.
The circuit is a series \(L\)-\(C\)-\(R\) combination fed by an AC source of rms voltage \(V\). The impedance of a series LCR circuit is
\[ Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \]
where \(X_{L}=\omega L\) is the inductive reactance and \(X_{C}=\dfrac{1}{\omega C}\) is the capacitive reactance.

Step 2: Apply the resonance condition.
At resonance the inductive and capacitive reactances become equal, so
\[ X_{L}=X_{C}\ \Rightarrow\ X_{L}-X_{C}=0 \]

Step 3: Find the impedance at resonance.
Substituting \(X_{L}-X_{C}=0\):
\[ Z=\sqrt{R^{2}+0}=R \]
So at resonance the circuit behaves as a pure resistance \(R\); the impedance is minimum.

Step 4: Apply Ohm's law for AC to get the current.
The same current flows through every element of a series circuit, so the current through \(R\) equals the source current:
\[ I=\frac{V}{Z}=\frac{V}{R} \]

Result: The current in resistance \(R\) at resonance is maximum and equals
\[\boxed{I=\dfrac{V}{R}}\]