Question

The ratio of the peak current through the capacitor and supply is known as –
A) Resonance current
B) Dynamic resistance
C) Q-factor
D) None of the above

Answer 1

We need to find a relation between the peak current through a capacitor and the current in the AC circuit to find the ratio between them. The equations for the given quantities- the peak current through the capacitor and the supply need to be found.

Complete answer:
We are to find the ratio of the peak current through the capacitor and the current through the supply and therefore, find the quantity that relates these quantities.
i.e.,

& \dfrac{{{I}_{c}}}{{{I}_{0}}}=\dfrac{{{V}_{0}}\omega C}{{{V}_{0}}\dfrac{CR}{L}} \\\ & \Rightarrow \dfrac{{{I}_{C}}}{{{I}_{0}}}=\dfrac{\omega L}{R}\text{ --(1)} \\\ \end{aligned}$$ Let us compare each of the given quantities to find the required answer. The resonance current is the current that resonates in the given AC circuit. It is the quantity that relates the resistive elements and reactive elements (inductance and capacitive elements). It is given by – $${{I}_{m}}=\dfrac{{{V}_{0}}}{\sqrt{{{R}^{2}}+{{({{X}_{c}}-{{X}_{L}})}^{2}}}}$$ The dynamic resistance is the resistance measured at any instant of time. It is like the instantaneous resistance for a given voltage and the corresponding current at a given instant of time. It is given as – $$r=\dfrac{\Delta v}{\Delta i}$$ The Q-factor is the dimensional quantity that relates the power input and power dissipated in a circuit. It determines the quality of a coil in an AC circuit. It relates the bandwidth to the frequency of the AC signal. It is defined as – $$Q=\dfrac{{{\omega }_{0}}}{2\Delta \omega }=\dfrac{{{\omega }_{0}}L}{R}$$ From this we understand that (1) is the same as the above equation. Therefore, the ratio of the peak current through the capacitor and supply is the Q-factor. **So, the correct answer is “Option C”.** **Note:** The Q-factor is a quantity that gives the idea about the bandwidth of the signal and the frequency of the signal. The greater the Q-factor, the bandwidth will be lesser and therefore, the resonance will be sharp, this is a required condition for making the signal better.