Answer to Question #216124 in Electricity and Magnetism for DANI

Question

For discharging process

R=10KiloOHM,V=470microF, E=12V.

A)What value will be the voltage across the capacitor 2 time constant

B)what value will be the voltage across after 6 seconds.

C)when the capacitor will be fully charged

D)when will the current through the circuit be reduced by 25%

Answer 1

A) Time constant

$τ=RC \\ = 10 \;k \Omega\times 470\; \mu F \\ = 4.7 \;s \\ t = 2 \times τ \\ t=9.4 \;s$

Voltage across the capacitor during charging

$V=V_0(1 -e^{-t/RC}) \\ = 12(1 -e^{-9.4/4.7}) \\ = 10.375\;V$

B) Voltage across the capacitor after 6 s

$V=V_0(1 -e^{-t/RC}) \\ = 12(1 -e^{-6/4.7}) \\ = 8.652 \;V$

C) At time t=4τ, voltage V=11.78 V

At time t=5τ, voltage V=11.92 V

At time t=6τ, voltage V=11.97 V

The choice is up to you, which time you want to consider because all time voltage is approximately equal to supply voltage.

D) Current in circuit

$I_c= \frac{V}{R}e^{-t/RC}$

When current reduced by 25%, remaining current 75%:

$0.75 = e^{-t/4.7} \\ ln(0.75)=-t/4.7 \\ t = 1.35 \;s$