question_answer

When AC circuit with L, C and R in series if brought into resonance, the current has large value. Why? If the capacitance C is increased, will the current increase or decrease? Explain with a suitable relation.

Answer:

The impedance of a series L-C-R circuit is given by                         \[Z=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\] When the circuit is brought into resonance, \[{{X}_{L}}={{X}_{C}}\] \[\therefore \]      \[\omega L=\frac{1}{\omega C}\] and Z = R Thus, the impedance Z of the circuit is minimum at resonance and hence, the current through the circuit is maximum. If capacitance C is increased, then reactance \[1/\omega C\] decreases, so impedance Z increase. Consequently, the current decreases.