Question

A capacitor of 2mF is charged with a DC Voltage source of 100 V . There is a resistor of 1 kilo ohms in series with the capacitor. What will be the time taken by the capacitor so that the voltage across the capacitor is 70 V

Answer 1

t = 0.731s

Step-by-step explanation:

In order to calculate the time that the capacitor takes to have a voltage of 70V, you use the following formula:

`V=V_oe^{-(t)/(RC)}` (1)

V: final voltage across the capacitor = 70V

Vo: initial voltage across the capacitor = 100V

R: resistance of the resistor in the circuit = 1kΩ = 1*10^3Ω

C: capacitance of the capacitor = 2mF = 2*10^-3F

t: time

You use properties of logarithms to solve the equation (1) for t:

`(V)/(V_o)=e^{-(t)/(RC)}\\\\ln((V)/(V_o))=ln(e^{-(t)/(RC)})\\\\ln((V)/(V_o))=-(t)/(RC)\\\\t=-RCln((V)/(V_o))`

Next, you replace the values of the parameters:

`t=-(1*10^3\Omega)(2*10^(-3))ln((70V)/(100V))\\\\t=0.713s`

The capacitor takes 0.731s to reache a voltage of 70V