Question

If reading of voltmeter V shown in the figure at resonance is 200 V, then the quality factor of the circuit is:

A- 1
B- 2
C- 3
D- 4

Answer 1

In the circuit diagram we have an AC source and there is a resistance and an inductor in series. Also, there is a capacitor and all of them are in series, so this is a series LCR circuit. The LCR circuit is at resonance, so we will use the condition of resonance.

Complete Step by step answer: At resonance the capacitive reactance is equal to the inductive reactance and as such the circuit behaves as purely resistive.
Voltage across the capacitor is 200 V. At resonance, voltage across the capacitor is equal to voltage across the inductor. Thus, voltage across the inductor is also 200 V.
At resonance, the voltage across capacitor and the source voltage and the quality factor are related as,

$${{V}_{C}}=Q\times {{V}_{S}} \\\ \therefore 200=Q\times 50 \\\ \Rightarrow Q=4 \\\$$

So, the quality factor comes out to be 4.

So, the correct option is (D).

Additional information: A voltmeter is a device which is used in electronics and to measure some value in electric circuits. The function that it performs is hidden in its name itself.
One of the most commonly used components that we encounter in physics labs is Rheostat. A rheostat is a variable resistor and is a box consisting of many resistances and is used in labs. The resistance which is taken out from it gets attached in to the circuit

Note: when we have a series LCR circuit, then the total opposition to the flow of the current is calculated by adding the resistance offered by the capacitor, the resistance and the inductor and that is known as impedance of the circuit. Quality factor related the energy stored in the circuit to the energy dissipated in each cycle of oscillation.