Question

6–3 The current through a 0.1‐μF capacitor is a rectangular pulse with an amplitude of 2 mA and a duration of 5 ms. Find the capacitor voltage at the end of the pulse when the capacitor voltage at the beginning of the pulse is –1 V.

Answer 1

999 V

Step-by-step explanation:

Charge deposited on the capacitor

= current x time

= 2 x 10⁻³ A x 5 x 10⁻³

= 10⁻⁵ C

increase in volt = charge deposited / capacitance

10⁻⁵ / .1 x 10⁻⁶

= 10² V

= 100 V

voltage in the beginning = -1 V

voltage at the end

= -1 + 100

99 V .