A parallel plate capacitor of capacitance 50 F is subjected to a time-varying voltage. The voltage across the capacitor increases uniformly from 0 to 10 V in the time interval t = 0 to t = 2 ms , and then remains constant at 10 V for t > 2 ms . The displacement current between the plates during the interval t = 0 to t = 2 ms is:

Step-by-step solution

The displacement current I_d is equal to the conduction current I_c during the charging of a capacitor, which is given by: I_d = C dV dt During the interval t = 0 to t = 2 ms , the voltage increases uniformly. The rate of change of voltage dV dt is the slope of the voltage-time graph: dV dt = V₂ - V₁ t₂ - t₁ dV dt = 10 - 0 2 10⁻³ - 0 = 10 2 10⁻³ = 5000 V s ⁻¹ Now, calculate the displacement current: I_d = (50 10⁻⁶ F ) (5000 V s ⁻¹) I_d = 250000 10⁻⁶ A = 250 10⁻³ A I_d = 250 mA Note: Using the voltage value directly