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QPages 330-331: Q16: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q16 Solution: An inductance of 2.0 H, a capacitance of 18 _F and a resistance …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly calculate the maximum current by including the inductive or capacitive reactance in the impedance formula, forgetting that at resonance $X_L = X_C$ and they cancel out, leaving the impedance equal to the pure resistance $R$. Always set the net reactance to zero before calculating $I_{max}$.

An inductance of 2.0 H, a capacitance of 18 _F and a resistance of 10 kΩ10 \text{ k}\Omega is connected to an AC source of 20 V with adjustable frequency. a) What frequency should be chosen to maximize the current in the circuit? b) What is the value of this maximum current?

Step-by-Step Verified Solution

**Answer:** (a) f≈27 Hzf \approx 27 \, \text{Hz} (b) Imax=2 mAI_{max} = 2 \, \text{mA}. **Step-by-Step Explanation:** (a) Resonance frequency ω=1LC\omega = \frac{1}{\sqrt{LC}}. L=2 HL = 2 \, \text{H}, C=18×10−6 FC = 18 \times 10^{-6} \, \text{F}. ω=136×10−6=16×10−3=166.67 rad/s\omega = \frac{1}{\sqrt{36 \times 10^{-6}}} = \frac{1}{6 \times 10^{-3}} = 166.67 \, \text{rad/s}. f=ω2π≈26.5 Hzf = \frac{\omega}{2\pi} \approx 26.5 \, \text{Hz}. (b) Imax=VR=20104=2 mAI_{max} = \frac{V}{R} = \frac{20}{10^4} = 2 \, \text{mA} [citation:4].

All Questions in this Chapter

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