An oscillator providing an output voltage E = 10 V t is connected to a capacitor of capacitance 10 F . Determine the peak currents in the circuit for angular frequencies = 10 s ⁻¹ , 100 s ⁻¹ , 500 s ⁻¹ , and 1000 s ⁻¹ .

Step-by-step solution

The peak current in a purely capacitive circuit is given by I₀ = V₀ X_C , where X_C = 1 C is the capacitive reactance. Substituting X_C , we get I₀ = V₀ C . Given V₀ = 10 V and C = 10 F = 10⁻⁵ F , the peak current is I₀ = 10 10⁻⁵ = 10⁻⁴ . For = 10 s ⁻¹ , I₀ = 10⁻⁴ 10 = 1.0 10⁻³ A . For = 100 s ⁻¹ , I₀ = 10⁻⁴ 100 = 0.01 A . For = 500 s ⁻¹ , I₀ = 10⁻⁴ 500 = 0.05 A . For = 1000 s ⁻¹ , I₀ = 10⁻⁴ 1000 = 0.1 A .