Answer:
1.) Vc = 1V
2.) Vc = 2.7V
3.) Time constant = 0.03
4.) V = 2.53V
Explanation:
1.) The value of Vc (in volts) just prior to the switch closing
The starting current = 1mA
With resistance R1 = 1000 ohms
By using ohms law
V = IR
Vc = 1 × 10^-3 × 1000
Vc = 1 volt.
2.) The value of Vc after the switch has been closed for a long time.
R2 and R3 are in parallel to each other. Both will be in series with R1
The equivalent resistance R will be
R = (R2 × R3)/R2R3 + R1
Where
R1 = 1000Ω,
R2 = 3000Ω,
R3 = 4000Ω
R = (4000×3000)/(4000+3000) + 1000
R = 12000000/7000 + 1000
R = 1714.3 + 1000
R = 2714.3 ohms
By using ohms law again
V = IR
Vc = 1 × 10^-3 × 2714.3
Vc = 2.7 volts
3.) The time constant = CR
Time constant = 10 × 10^-6 × 2714.3
Time constant = 0.027
Time constant = 0.03 approximately
4.) The value of Vc at t = 2msec (in volts). Can be calculated by using the formula
V = Vce^-t/CR
Where
Vc = 2.7v
t = 2msec
CR = 0.03
Substitute all the parameters into the formula
V = 2.7 × e^-( 2×10^-3/0.03)
V = 2.7 × e^-(0.0667)
V = 2.7 × 0.935
V = 2.53 volts
