Question

Three students, X, Y and Z performed an experiment for studying the variation of ac with frequency in a series LCR circuit and obtained the graphs as shown below. They all used

• an AC source of the same emf and
• inductance of the same value.

1. Who used minimum resistance?
2. In which case will the quality Q factor be maximum?
3. What did the students conclude about the nature of impedance at resonant frequency (f₀)?
4. An ideal capacitor is connected across 220 V, 50 Hz, and 220 V, 100 Hz supplies. Find the ratio of current flowing through it in the two cases.

Answer 1

a. At resonance frequency f₀​, the impedance of a series LCR circuit is minimum and equals R.

I₀ = `V/Z = V/R`

Hence, greater peak current implies smaller resistance. From the graph, curve X has the maximum peak current; therefore, X used the minimum resistance (and Z used the maximum resistance).

b. The quality factor of a series LCR circuit is given by

Q = `f_0/(Δf)`

where Δf is the bandwidth. A sharper (narrower) resonance curve means smaller bandwidth and hence larger Q. From the graph, curve X is the sharpest; therefore, the quality factor is maximum for X.

c. At resonance frequency f₀, the inductive reactance equals the capacitive reactance (X_L = X_C). Hence, the net reactance is zero and the impedance of the circuit is minimum and purely resistive, i.e., Z = R.

d. For a capacitor,

`X_C = 1/(2 pi fC)`

⇒ I = `V/(X_C)`

⇒ I = V (2πfC)

Hence, the current is directly proportional to frequency (I ∝ f).

Therefore, for 50 Hz and 100 Hz,

I₅₀ : I₁₀₀

= 50 : 100

= 1 : 2