Question

In a series resonant LCR circuit, the voltage across \(R\) is \(100 \mathrm{~V}\) and \(\mathrm{R}=1 \mathrm{k} \Omega\) with \(\mathrm{C}=2 \mu \mathrm{F}\). The resonant frequency 0 is \(200 \mathrm{rad} / \mathrm{s}\). At resonance the voltage across \(\mathrm{L}\) is. (a) \(40 \mathrm{~V}\) (b) \(250 \mathrm{~V}\) (c) \(4 \times 10^{-3} \mathrm{~V}\) (d) \(2.5 \times 10^{-2} \mathrm{~V}\)

Short answer

The voltage across the inductor (L) at resonance is 250 V. Therefore, the correct answer is (b) 250 V.

Step-by-step solution

1. Calculate the current in the circuit at resonance

: At resonance, the impedances of the inductor and capacitor are equal, so the total impedance of the circuit is just the resistance. The voltage across the resistor is given as 100 V, and the resistance is given as 1 kΩ. We can use Ohm's law to find the current (I) in the circuit. \(I = \frac{V_R}{R}\)

2. Plug in the given values to calculate the current

: Now, plug in the values for the voltage across R (100 V) and the resistance R (1 kΩ) into the equation we obtained in step 1. \(I = \frac{100 \mathrm{V}}{1 \mathrm{k\Omega}}\) \(I = 0.1 \mathrm{A}\)

3. Calculate the impedance of the inductor (XL)

: The impedance of an inductor is given by: \(X_L = \omega L\) where ω is the resonant frequency and L is the inductance. We are given the resonant frequency, ω = 200 rad/s, but we need to rearrange the equation to find the inductance L. \(L=\frac{X_L}{\omega}\)

4. Calculate the impedance of the capacitor (XC)

: The impedance of a capacitor is given by: \(X_C = \frac{1}{\omega C}\) Plug in the given values for ω (200 rad/s) and C (2 μF) to calculate XC. \(X_C = \frac{1}{(200 \mathrm{rad/s})(2 \times 10^{-6} \mathrm{F})}\) \(X_C = 2.5 \mathrm{k\Omega}\)

5. Use the current and XC to find the voltage across the inductor (VL)

: At resonance, the impedances of the inductor and capacitor are equal (XL = XC). Therefore, we can use either the impedance of the inductor or the capacitor to calculate the voltage across the inductor (VL) using Ohm's law. \(V_L = IX_C\) Plug in the values for the current I (0.1 A) and XC (2.5 kΩ) into the equation obtained in step 5. \(V_L = (0.1 \mathrm{A})(2.5 \mathrm{k\Omega})\) \(V_L = 250 \mathrm{V}\) The voltage across the inductor (L) at resonance is 250 V. Hence, the correct answer is: (b) 250 V