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.

999 V

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 .

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s