For the circuit, suppose C=10µF, R1=1000Ω, R2=3000Ω, R3=4000Ω and ls=1mA. The switch closes at t=0s.1) What is the value of Vc (in volts) just prior to the switch closing? Assume that the switch had been open for a long time. 2) For the circuit above, what is the value of Vc after the switch has been closed for a long time? 3) What is the time constant of the circuit (in seconds)? Enter the answer below without units.
4) What is the value of Vc at t = 2msec (in volts).

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

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be