Physics · Alternating Current · NEET
No. The formula omega_0 = 1/sqrt(LC) contains only L and C. R decides how sharp the resonance peak is (the Q-factor and bandwidth) and how large the maximum current is (i_max = V/R), but it never shifts the resonant frequency. This is a favourite NEET trap: a value of R is given only to distract you.
omega_0 is the resonant ANGULAR frequency in rad/s: omega_0 = 1/sqrt(LC). f_0 is the ordinary resonant frequency in hertz (cycles per second): f_0 = omega_0/(2*pi) = 1/(2*pi*sqrt(LC)). If the source is written as V = V_0*sin(omega*t), the number multiplying t is omega (rad/s), not f. Always check whether the options are in rad/s or Hz.
The impedance is Z = sqrt(R^2 + (X_L - X_C)^2). When X_L = X_C, the reactive part (X_L - X_C) becomes zero, so Z drops to its smallest value Z = R. With Z minimum, the current amplitude i = V/Z is maximum. That peak-current condition IS resonance.
Start from the resonance condition X_L = X_C. Write omega_0*L = 1/(omega_0*C). Multiply both sides by omega_0: omega_0^2 * L = 1/C. So omega_0^2 = 1/(LC), giving omega_0 = 1/sqrt(LC). Divide by 2*pi to get f_0 = 1/(2*pi*sqrt(LC)). Three lines, full marks.
Yes. Always convert: mH to H (multiply by 10^-3), micro-F to F (multiply by 10^-6), nF to F (10^-9). A common mistake is leaving C in micro-F, which makes the answer wrong by a factor of 1000. Compute the product LC in SI units before taking the square root.
A series LCR circuit with inductance 10 H, capacitance 10 microF, 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:
In a series LCR circuit, the inductance L is 10 mH, capacitance C is 1 microF and resistance R is 100 ohm. The frequency at which resonance occurs is:
An ac circuit contains a resistance of 1 kohm, a capacitor of 0.1 microF and an inductor of 1 mH connected in series. The resonance frequency of the circuit is approximately:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The resonant angular frequency is omega_0 = 1/sqrt(LC) in rad/s, and the resonant frequency is f_0 = 1/(2*pi*sqrt(LC)) in hertz. L is in henry and C in farad.
Resonance is set by the balance of inductive and capacitive reactance (X_L = X_C), and only L and C appear in those reactances. R affects the sharpness (Q-factor) and the peak current, but not the frequency at which resonance occurs.
Impedance is minimum, Z = R, because the reactive part (X_L - X_C) becomes zero. The current is therefore maximum, i_max = V/R, and the circuit behaves like a pure resistor with power factor 1.
The condition X_L = X_C giving omega_0 = 1/sqrt(LC) is the same. But at resonance a series LCR gives MINIMUM impedance and maximum current, while an ideal parallel LCR gives MAXIMUM impedance and minimum current. NEET usually tests the series case.
Convert L to henry and C to farad first (mH x10^-3, microF x10^-6, nF x10^-9), multiply to get LC, take the square root, then apply 1/(2*pi*sqrt(LC)). Check whether the answer is asked in rad/s (omega_0) or Hz (f_0).
