LR, RC and LCR Circuits: Practice Problem & Solution
A series LCR circuit with inductance $10 H$, capacitance $10 \mu F$, resistance $50 \Omega$ is connected to an ac source of voltage, $V = 200 \sin(100t)$ volt. If resonant frequency of the LCR circuit is $\nu_0$ and the frequency of the ac source is $\nu$, then: (2022)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of LR, RC and LCR Circuits. Understanding the underlying formula is key to arriving at the correct answer below:
For the AC source $V = 200 \sin(100t)$, angular frequency $\omega = 100 rad/s$.
Source frequency $\nu = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi} Hz$.
Resonant frequency $\nu_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10 \times 10 \times 10^{-6}}} = \frac{1}{2\pi\times 10^{-2}} = \frac{50}{\pi} Hz$.
Thus, $\nu_0 = \nu = \frac{50}{\pi} Hz$.
Detailed Solution (Best of Understanding Complex Concepts)
Concepts Involved
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AC Source Frequency: The voltage of an alternating current source is given by the standard equation $V = V_0 \sin(\omega t)$, where $V_0$ is the peak voltage and $\omega$ is the angular frequency. The linear source frequency $\nu$ is related to angular frequency by the formula $\nu = \frac{\omega}{2\pi}$.
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Resonant Frequency in an LCR Circuit: Resonance in a series LCR circuit occurs when the inductive reactance ($X_L$) perfectly balances the capacitive reactance ($X_C$), minimizing impedance. The resonant frequency $\nu_0$ is calculated using the formula $\nu_0 = \frac{1}{2\pi\sqrt{LC}}$, where $L$ is the inductance and $C$ is the capacitance.
Step-by-Step Solution
Step 1: Identify the given values
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Inductance, $L = 10\text{ H}$
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Capacitance, $C = 10\mu\text{F} = 10 \times 10^{-6}\text{ F}$
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Resistance, $R = 50 \Omega $ (Not needed for frequency calculations)
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Voltage equation: $V = 200 \sin(100t)$
Step 2: Calculate the frequency of the AC source ($\nu$)
By comparing the given voltage equation $V = 200 \sin(100t)$ with the standard form $V = V_0 \sin(\omega t)$, we can extract the angular frequency:
$$\omega = 100\text{ rad/s}$$
Now, calculate the linear frequency $\nu$:
$$\nu = \frac{\omega}{2\pi}$$
$$\nu = \frac{100}{2\pi} = \frac{50}{\pi}\text{ Hz}$$
Step 3: Calculate the resonant frequency ($\nu_0$)
Using the standard formula for the resonant frequency of an LCR circuit:
$$\nu_0 = \frac{1}{2\pi\sqrt{LC}}$$
Substitute the given values for $L$ and $C$ into the equation:
$$\nu_0 = \frac{1}{2\pi\sqrt{10 \times (10 \times 10^{-6})}}$$
$$\nu_0 = \frac{1}{2\pi\sqrt{100 \times 10^{-6}}}$$
$$\nu_0 = \frac{1}{2\pi\sqrt{10^{-4}}}$$
$$\nu_0 = \frac{1}{2\pi \times 10^{-2}}$$
$$\nu_0 = \frac{100}{2\pi} = \frac{50}{\pi}\text{ Hz}$$
Final Answer
Both the source frequency and the resonant frequency are equal:
$$\nu_0 = \nu = \frac{50}{\pi}\text{ Hz}$$
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