Physics · Alternating Current · NEET
At resonance the inductive reactance and capacitive reactance become equal: XL = XC. Since they act in opposite directions in the phasor diagram, they cancel each other. The net reactance (XL - XC) becomes zero, so the impedance Z = sqrt(R^2 + (XL - XC)^2) shrinks to just Z = R. Because Z is now smallest, the current I = V/Z reaches its maximum value. The circuit behaves exactly like a pure resistor at this one frequency.
Current in an AC circuit is I = V / Z. The impedance Z is smallest when the reactive part (XL - XC) is zero, which is exactly the resonance condition. With Z = R (the minimum possible value), I = V / R is as large as it can be. So current peaks sharply at the resonant frequency and falls off on either side. This is why resonance is used to 'tune' radios to one station.
Impedance is Z = sqrt(R^2 + (XL - XC)^2). At resonance XL = XC, so the reactance term vanishes and Z = sqrt(R^2) = R. It cannot be zero because the resistor R is still present and never cancels — only the reactances cancel. So the minimum impedance equals R, not zero. A common trap is writing Z = 0 at resonance; the correct answer is Z = R.
No. The resonant frequency is set only by L and C: it occurs when XL = XC, giving omega0 = 1 / sqrt(LC). The resistance R does not appear in this condition. R controls how sharp the resonance peak is (the quality factor and bandwidth), but it does not shift where resonance happens. In PYQs you can ignore R entirely when finding the resonant frequency.
At resonance the circuit is purely resistive, so the current and voltage are in phase. The phase angle phi = 0, which means cos phi = 1 (power factor is unity). This is the maximum possible power factor, so the circuit delivers maximum average power to R at resonance. If a question tells you the power factor is 1 in an LCR circuit, that circuit is at resonance.
A series LCR circuit with inductance 10 H, capacitance 10 uF, resistance 50 ohm is connected to an ac source of voltage, V = 200 sin(100 t) volt. If the resonant frequency of the LCR circuit is v0 and the frequency of the ac source is v, then:
A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is pi/3. If instead C is removed, the phase difference is again pi/3. The power factor of the circuit is:
An ac voltage V = 220 sin(2 x 10^3 t) V is applied to a series LCR circuit. The current amplitude in the circuit is: [L = 10 mH, C = 25 uF, R = 100 ohm]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The condition is XL = XC, meaning inductive reactance equals capacitive reactance. This happens at the resonant angular frequency omega0 = 1 / sqrt(LC). At this frequency the net reactance is zero.
The impedance is minimum and equals the resistance: Z = R. Since Z = sqrt(R^2 + (XL - XC)^2) and XL = XC at resonance, the reactance term vanishes, leaving Z = R.
At resonance the impedance is minimum and the current is maximum, so the circuit 'accepts' the signal at the resonant frequency most strongly. This is why series resonance is used to tune (select) a particular radio or TV station frequency.
Yes. At resonance the circuit is purely resistive, current and voltage are in phase, the phase angle is 0, so power factor cos phi = 1. This gives maximum average power dissipated in the resistor.
No. Resonant frequency omega0 = 1 / sqrt(LC) depends only on L and C. Resistance R only affects the sharpness of the peak (quality factor and bandwidth), not the frequency at which resonance occurs.
