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QPages 330-331: Q8: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q8 Solution: A capacitor of capacitance 10 _F is connected to an oscillator…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often insert the given capacitance value (e.g., 10 µF) directly into $I_0=V_0\omega C$ without converting µF to farads, producing a current that is off by a factor of $10^6$. Always express $C$ in farads before applying $I_0=V_0\omega C$.

A capacitor of capacitance 10 _F is connected to an oscillator with output voltage ε=10 V)sin⁡ωt\varepsilon = 10 \text{ V}) \sin \omega t. Find the peak currents in the circuit for ω=10 s−1\omega = 10 \text{ s}^{-1}, 100 s−1100 \text{ s}^{-1}, 500 s−1500 \text{ s}^{-1} and 1000 s−11000 \text{ s}^{-1}.

Step-by-Step Verified Solution

**Answer:** 0.001,0.01,0.05,0.1 A0.001, 0.01, 0.05, 0.1 \, \text{A}. **Step-by-Step Explanation:** I0=E0XC=E0ωCI_0 = \frac{E_0}{X_C} = E_0 \omega C. E0=10 VE_0 = 10 \, \text{V}, C=10−5 FC = 10^{-5} \, \text{F}. For ω=10,100,500,1000 s−1\omega = 10, 100, 500, 1000 \, \text{s}^{-1}: I0=10×ω×10−5=ω×10−4 AI_0 = 10 \times \omega \times 10^{-5} = \omega \times 10^{-4} \, \text{A} [citation:6][citation:9].

All Questions in this Chapter

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