MP Board · Class 12 · Physics · Alternating CurrentAn alternating voltage $V = 200\sqrt{2} \sin(1000t)\text{ V}$ is applied across a series LCR circuit containing a resistor $R = 400\ \Omega$, an inductor $L = 0.1\text{ H}$, and a capacitor $C = 2.5\ \mu\text{F}$. \nCalculate the following step-by-step: (i) Inductive reactance ($XL$) and Capacitive reactance ($XC$) (ii) Total impedance ($Z$) of the circuit (iii) Peak current ($I0$) and RMS current ($I{\text{rms}}$) (iv) Power factor ($\cos\phi$) of the circuit (v) Resonant frequency ($fr$) of the given circuit

Step-by-Step Numerical Solution

Given Parameters:

• Supply voltage equation: $V(t) = 200\sqrt{2} \sin(1000t)\text{ V}$
• Comparing with standard AC voltage equation $V(t) = V_0 \sin(\omega t)$:
  • Peak Voltage ($V_0$) = $200\sqrt{2}\text{ V}$
  • Angular frequency ($\omega$) = $1000\text{ rad/s}$
  • RMS Voltage ($V_{\text{rms}}$) = $\frac{V_0}{\sqrt{2}} = \frac{200\sqrt{2}}{\sqrt{2}} = 200\text{ V}$
• Resistance ($R$) = $400\ \Omega$
• Inductance ($L$) = $0.1\text{ H}$
• Capacitance ($C$) = $2.5\ \mu\text{F} = 2.5 \times 10^{-6}\text{ F}$

Step (i): Calculation of Inductive Reactance ($X_L$) and Capacitive Reactance ($X_C$)

1. Inductive Reactance ($X_L$): $$X_L = \omega L$$ $$X_L = 1000 \times 0.1 = 100\ \Omega$$

2. Capacitive Reactance ($X_C$): $$X_C = \frac{1}{\omega C}$$ $$X_C = \frac{1}{1000 \times 2.5 \times 10^{-6}} = \frac{1}{2.5 \times 10^{-3}} = \frac{1000}{2.5} = 400\ \Omega$$

Step (ii): Calculation of Total Impedance ($Z$)\nThe total impedance formula for a series LCR circuit is:

$$Z = \sqrt{R^2 + (X_C - X_L)^2}$$ \nSubstituting values: $$Z = \sqrt{(400)^2 + (400 - 100)^2}$$ $$Z = \sqrt{400^2 + 300^2}$$ $$Z = \sqrt{160000 + 90000} = \sqrt{250000}$$ $$Z = 500\ \Omega$$

Step (iii): Calculation of Peak Current ($I_0$) and RMS Current ($I_{\text{rms}}$)

1. Peak Current ($I_0$): $$I_0 = \frac{V_0}{Z} = \frac{200\sqrt{2}}{500} = 0.4\sqrt{2}\text{ A} \approx 0.566\text{ A}$$

2. RMS Current ($I_{\text{rms}}$): $$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200}{500} = 0.4\text{ A}$$

Step (iv): Calculation of Power Factor ($\cos\phi$)\nPower factor is given by:

$$\cos\phi = \frac{R}{Z}$$ $$\cos\phi = \frac{400}{500} = 0.8$$ (Since $X_C > X_L$, the circuit is capacitive and current leads the voltage).

Step (v): Calculation of Resonant Frequency ($f_r$)\nThe formula for resonant frequency is:

$$f_r = \frac{1}{2\pi \sqrt{LC}}$$ \nSubstituting $L$ and $C$: $$\sqrt{LC} = \sqrt{0.1 \times 2.5 \times 10^{-6}} = \sqrt{0.25 \times 10^{-6}} = 0.5 \times 10^{-3}\text{ s}$$ $$f_r = \frac{1}{2\pi \times (0.5 \times 10^{-3})} = \frac{1000}{\pi}\text{ Hz} \approx \frac{1000}{3.1416} \approx 318.31\text{ Hz}$$