A series LCR circuit connected to an AC source with voltage of the source $v=v_m \sin \omega t$.
If 'q' is the charge on the capacitor and 'i' is the current, from Kirchoff's loop rule:
$L \frac{d i}{d t}+i R+\frac{q}{c}=V$
The current in the circuit is given by $i=I_{m} \sin (\omega t+ \phi)$ where $\phi$ is the phase difference between the voltage across the source and current in the circuit.
We know $V_{R m}=L_m R ; V_{L m}=L_m X_L ; V_{C m}=L_m X_C$; and $X_L=\omega L ; X_C=\frac{1}{\omega C}$
Total impedance in the circuit regulates current. At resonance frequency of the LCR circuit current in the circuit is maximum.
A pure inductor of 0.25 mH, a pure capacitor of 250 µF and a resistor of $350 \Omega$ are connected in series. An ac source of amplitude 210 V is connected across this series combination of L, C and R. The impedance and current amplitude in the circuit at resonance will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $Z_{\text {min }}=350 \Omega$ and $I_{\max }=0.60 ~A$
Impedance of the circuit, Z = R [at resonance]
$\Rightarrow Z=R=350 \Omega$
At resonance impedance is minimum, $\Rightarrow Z_{\text {min }} =350 \Omega$
$I_{\max }=\frac{V_{\max }}{R} \Rightarrow I_{\max }=\frac{210}{350}=0.60 ~A$