Typical Numerical’s on LCR Circuit - Applications

JEE( Main):

1. Given, L = 100 H, C = 100 μF, R = 100 Ω, Vrms = 100 V, f = 50 Hz

a. The resonant frequency (f0) of the circuit is given by:

   $$f_0 = \frac{1}{2π√(LC)}$$

  $$f_0 = \frac{1}{2π√(100 × 10^{-2} × 100 × 10^{-6})}$$

  $$f_0 $$ = 15.92 Hz

b. The impedance (Z) of the circuit at resonance is equal to the resistance (R):

   $$Z = R = 100 Ω$$

c. The current (I) flowing through the circuit at resonance is given by:

   $$I = \frac{V_{rms}}{Z}$$

   $$I = \frac{100V}{100 Ω}$$

   $$I $$ = 1 A

2. Given, N_P = 100 turns, N_S = 200 turns, Vrms = 230 V, f = 50 Hz

a. The voltage (Vrms_S) across the secondary coil is:

     $$V_{rms_S} = \frac{N_S}{N_p} × Vrms_p$$

     $$V_{rms_S} = \frac{200 turns}{100 turns} × 230V $$

     $$V_{rms_S}$$ = 460 V

b. The current (I_S) flowing through the secondary coil is:

    $$I_S = \frac{Vrms_S}{R_L}$$

    $$I_S = \frac{460V}{100Ω}$$

    $$I_S $$= 4.6 A

c. The power (P) transferred to the load is given by:

    $$P = I_S^2 × R_L$$

   $$P = (4.6A)^2 × 100Ω$$

   $$P $$= 2116 W

3. Given, R = 20Ω, L = 0.1 H, C = 100μF, Vrms = 100V, f = 50 Hz

a. The resonant frequency (f0) is:

   $$f_0 = \frac{1}{2π√(LC)}$$

   $$f_0 = \frac{1}{2π√(0.1 × 100 × 10^{-6})}$$

   $$f_0 =$$  159.2 Hz

b. The impedance (Z) of the circuit at resonance is given by:

    $$Z = R = 20 Ω$$

c. The current (I) flowing through the circuit at resonance is:

   $$I = \frac{Vrms}{Z}$$

   $$I = \frac{100V}{20Ω}$$

   $$I =$$ 5 A

CBSE (Board):

1. Given, L = 50 H, C = 50 μF, R = 50 Ω, Vrms = 50 V, f = 50 Hz

a. The resonant frequency (f0) is:

   $$f_0 = \frac{1}{2π√(LC)}$$

   $$f_0 = \frac{1}{2π(50 × 10^{-2} × 50 × 10^{-6})}$$

   $$f_0 =$$  159.2 Hz

b. The impedance (Z) of the circuit at resonance is equal to the resistance (R):

   $$Z = R = 50 Ω$$

c. The current (I) flowing through the circuit at resonance is:

   $$I = \frac{Vrms}{Z}$$

   $$I = \frac{50V}{50Ω}$$

   $$I =$$ 1 A

2. Given, N_P = 100 turns, N_S = 200 turns, Vrms = 230 V, f = 50 Hz

a. The voltage (Vrms_S) across the secondary coil is:

     $$V_{rms_S} = \frac{N_S}{N_p}×Vrms_p$$

     $$V_{rms_S}=\frac{200}{100}×230$$

     $$V_{rms_S}=460$$ V

b. The current (I_S) flowing through the secondary coil is:

   $$I_S = \frac{Vrms_S}{R_L}$$

   $$I_S = \frac{460V}{50Ω}$$

   $$I_S =$$  9.2 A

c. The power (P) transferred to the load is given by:

      $$P = I_S^2 × R_L$$

      $$P = (9.2A)^2 × 50Ω$$

      $$P =$$  4232 W

3. Given, R = 10 Ω, L = 0.05 H, C = 50 μF , Vrms = 50 V, f = 50 Hz

a. The resonant frequency (f0) is:

    $$f_0 = \frac{1}{2π√(LC)}$$

    $$f_0 = \frac{1}{2π√(0.05 × 50 × 10^{-6})}$$

    $$f_0 =$$ 223.6 Hz

b. The impedance (Z) of the circuit at resonance is given by:

     $$Z = R = 10 Ω$$

c. The current (I) flowing through the circuit at resonance is:

    $$I = \frac{Vrms}{Z}$$

    $$I = \frac{50V}{10Ω}$$

    $$I =$$ 5 A